Method for checking the suitability of a target trajectory for trajectory control of a vehicle

ABSTRACT

A method for checking the suitability of a target trajectory for trajectory control of a vehicle. The target trajectory contains route information about the path of a route to be driven and momentum information about the momentum with which the route should be driven. The target trajectory is assessed as suitable for the trajectory control if the target traction and target tractive power are below a predetermined characteristic traction curve or characteristic tractive power curve, the target steering power is below a predetermined power limit of the steering, and the horizontal target tire forces on each wheel are within the friction value limits determined by the friction value. Otherwise, the target trajectory is assessed to be unsuitable.

BACKGROUND AND SUMMARY OF THE INVENTION

Exemplary embodiments of the invention relate to a method for checkingthe suitability of a target trajectory for trajectory control of avehicle.

In manually driven vehicles, the driver assumes the task of planning thetrajectory, i.e., they determine on which route they would like totravel when and at what speed. In planning the trajectory, the drivertakes into consideration the characteristics of the road surface infront of the vehicle as well as the acquired knowledge about the vehiclereactions to expect. With this information, the driver plans atrajectory that the vehicle will, according to experience, follow, i.e.,the trajectory can be travelled.

DE 100 50 421 A1 discloses a road performance control method of an, inparticular, four-wheeled motor vehicle. The potential traction that isavailable between the wheels and the road surface is detected and isdetermined by means of a comparison with the likewise detected tractionof a traction reserve that is currently being exploited. According tothe invention, besides the forces acting in the horizontal plane, thevertical movements of the motor vehicle structure compared to the wheelsare additionally taken into consideration during the determining of thetraction reserve. Preferably, the maximum horizontal force that can betransferred by the wheel tires is detected by means of multiplying thewheel load orientated in the vertical direction with a value of thefriction between the wheel tires and the road surface, which isestimated or detected by means of a sensor, from which horizontal forcethe traction reserves in the longitudinal direction of the vehicle andin the transverse direction of the vehicle and in the vertical directionare detected, with the suitably estimated current values for thelongitudinal force and lateral force acting in the horizontal plane. Thedetected traction reserves can then be communicated to a so-calledtraction controller, which arranges a favorable use of the currentavailable traction, taking into consideration the desired drivingmaneuver.

Exemplary embodiments of the invention are directed to an improvedmethod for checking the suitability of a target trajectory fortrajectory control of a vehicle.

In a method according to the invention for checking the suitability of atarget trajectory for trajectory control of a vehicle, the targettrajectory contains route information about the path of a route to bedriven and momentum information about the momentum with which the routeshould be driven. According to the invention, a slope, a transverseincline, and a friction value of a road surface along the targettrajectory are detected and/or estimated by sensors and/or from mapdata. If some values cannot be detected by sensors, these are, forexample, taken from map data. If the values cannot be either directlymeasured, nor taken directly from map data, then the values areestimated, for example based on sensor information or map data. Targetvalues, required for driving the target trajectory, of target traction,target tractive power, target steering power, and horizontal target tireforces on individual wheels of the vehicle are calculated based on modelequations of the vehicle from the route and momentum information aboutthe target trajectory and from the detected slope and transverseincline, wherein the target trajectory is assessed as suitable for thetrajectory control if:

-   the target traction and target tractive power are below a    predetermined characteristic traction curve or characteristic    tractive power curve,-   the target steering power is below a predetermined power limit of    the steering and-   the horizontal target tire forces on each wheel are within the    friction value limits determined by the friction value,-   wherein the target trajectory is otherwise assessed to be    unsuitable.

In an embodiment, the method is used on every target trajectory from apredetermined plurality of target trajectories.

In an embodiment, every target trajectory assessed as unsuitable isrejected as invalid.

In an embodiment, the characteristic traction curve and thecharacteristic tractive power curve are taken from respective look-uptables.

In an embodiment, the characteristic traction curve and thecharacteristic tractive power curve take into consideration occurrencesof degradation of a drive train.

In an embodiment, a quantitative assessment of the exploitation of thepotential friction value is carried out.

In an embodiment, occurrences of degradation of a steering actuator aretaken into consideration.

In an embodiment, the model equations of the vehicle rely on aquasi-steady-state modelling approach.

In an embodiment, in a first step, a conversion of parameters of thetarget trajectory as well as of the slope and transverse incline intovehicle parameters is carried out by means of the model equations.

In an embodiment, the target traction and the target tractive power fora center of gravity of the vehicle are calculated.

The proposed invention solves the problem of determining drivability forautonomous vehicles.

The invention relates to a method for checking the suitability of atarget trajectory for trajectory control of a vehicle, wherein thetarget trajectory contains route information about the path of a routeto be driven (target position, target camber, target change in camber)and momentum information about the momentum (target speed, targetacceleration) with which the route should be driven.

According to the invention, the slope λ, the transverse incline η, andthe friction value µ_(max) of the road surface along the targettrajectory are detected or estimated (by sensors or from map data).Furthermore, the target values, required for driving the targettrajectory, of the driving force (target traction F_(traction,demand)),of the driving power (target tractive power P_(traction,demand)), of thesteering power (target steering power P_(steering,max)), and thehorizontal target tire forces on the individual wheels (longitudinaltarget tire forces F_(XT,i), transverse target tire forces F_(YT,i)) arecalculated based on model equations of the vehicle from the route andmomentum information about the target trajectory and from the detectedslope λ and transverse incline η. The target trajectory is assessed assuitable for the trajectory control if a test shows that

-   the target traction and target tractive power are below a    predetermined characteristic traction curve or characteristic    tractive power curve,-   the target steering power is below a predetermined power limit of    the steering and-   the horizontal target tire forces on each wheel are within the    friction value limits determined by the friction value µ_(max)    (maximum available potential traction, traction ellipse).

Otherwise, the target trajectory is assessed as unsuitable for thetrajectory control. Advantageously, the method is used on every targettrajectory from a predetermined plurality of target trajectories. Everytarget trajectory assessed as unsuitable is rejected as invalid. Thecharacteristic traction curve and the characteristic tractive powercurve are advantageously taken from a look-up table and take intoconsideration possible occurrences of degradation.

The invention further relates to a device that is configured to carryout a method as described above. In particular, the device can comprisea data processing device, for example a control unit in a motor vehicle.

BRIEF DESCRIPTION OF THE DRAWING FIGURES

Exemplary embodiments of the invention are described in more detail inthe following, using drawings.

Here:

FIG. 1 shows a schematic side view of a vehicle with different axlesystems,

FIG. 2 shows a schematic rear view of a vehicle with different axlesystems,

FIG. 3 shows a schematic plan view of a vehicle with different axlesystems,

FIG. 4 shows schematic rotations of axle systems,

FIG. 5 shows a schematic view of the vehicle, wherein the roll axis, thefront roll center and the rear roll center are shown,

FIG. 6 shows a schematic view of the vehicle,

FIG. 7 shows a schematic view of the vehicle with front and rear axialforces and torques,

FIG. 8 shows a schematic view of the front axle with tire forces,chassis reaction forces and the chassis reaction torque,

FIG. 9 shows schematic views of look-up-tables,

FIG. 10 shows a schematic view of a tire forces-ellipse, and

FIG. 11 shows a schematic close-up view of a functional design of adevice for model-based testing of drivability.

Parts that correspond to each other are labelled with the same referencenumerals in all figures.

DETAILED DESCRIPTION

The present invention relates to the assessing of the drivability of alldesired trajectories from a number of potential trajectories that aregiven in fixed coordinates as horizontal trajectories, whereindrivability limits for trajectories are typically given as vehicle androad surface parameters. Examples are the potential friction, whichlimits the maximum attainable horizontal tire forces, and the drivingforces and power limits in the drive train, which limit possibleacceleration. For testing these limits, it is necessary to calculate theassociated vehicle parameters from the desired trajectories. Herein, aquasi-steady-state (QSS) modelling approach is selected, in order tocalculate both tire forces and also the driving force and power limitsfrom a given, desired trajectory. Herein, an amount of, for example,1000 desired possible trajectories with a length of 10 s was used, whichwere evaluated every 100 ms. The running time is therefore an importantpoint and the main reason for the choice of a quasi-steady-statemodelling approach instead of, for example, a dynamic model, whichrequires more exact sampling and solving of the common differentialequations of this model.

The axle systems used in the following obey ISO 8855:2011. The mostimportant axle systems are indicated below. A graphic overview ofrelevant perpendicular axle systems, kinematic characteristics, forcesand torques is shown in FIGS. 1, 2 and 3 .

The vehicle axle system (X_(V), Y_(V), Z_(V)) is an axle system that isrooted in the reference frame of the sprung vehicle mass so that theX_(V) axis is essentially aligned horizontally and to the front (if thevehicle 1 is at a standstill) and parallel to the longitudinal plane ofsymmetry of the vehicle 1. The Y_(V) axis is perpendicular to thelongitudinal plane of symmetry of the vehicle 1 and points to the left,wherein the Z_(V) axis points upwards. The source of the associatedvehicle coordinate system (x_(V), y_(V), z_(V)) is in the middle of thefront axle during static reference load conditions.

The fixed axle system (X_(E), Y_(E), Z_(E)) is an axle system that isfixed in the context of the vehicle being stationary. X_(E) and Y_(E)are parallel to the plane of the ground. Z_(E) points upwards and isaligned with the gravity vector. The source of the associated fixedcoordinate system (X_(E), Y_(E), Z_(E)) is on the plane of the ground.

The interconnected, levelled axle system (X, Y, Z) is an axle system,the X and Y axes of which are parallel to the plane of the ground,wherein the X axis is aligned on the vertical projection of the X_(V)axis, on the plane of the ground. The source of the associatedcoordinate system (x, y, z) coincides with the source of the vehiclecoordinate system.

The road surface plane axle system (X_(R), Y_(R), Z_(R)) is an axlesystem, the X_(R) and Y_(R) axes of which are parallel to the plane ofthe road surface, wherein the X_(R) axis is aligned on the verticalprojection of the X_(V) axis, on the road surface plane. The source ofthe associated road surface coordinate system (X_(R), Y_(R), Z_(R))coincides with the source of the vehicle axle system. The road surfaceplane is a best-fit for the four contact points of the tires.

The tire axle system (X_(T,1R), Y_(T,1R), Z_(T,1R)) for the right frontwheel 1R, the tire axle system (X_(T,1L), Y_(T,1L), Z_(T,1L)) for theleft front wheel 1L, the tire axle system (X_(T,2R), Y_(T,2R), Z_(T,2R))for the right rear wheel 2R and the tire axle system (X_(T,2L),Y_(T,2L), Z_(T,2L)) for the left rear wheel 2L are axle systems, theX_(T) and Y_(T) axes of which are parallel to the plane of the roadsurface, wherein the Z_(T) axis is aligned normal to the road surfaceplane, wherein the alignment of the X_(T) axis is defined by theintersection of the wheel plane and the road surface plane, wherein thepositive Z_(T) axis points upwards. FIG. 1 further shows the angle ofincline θ (negatively shown) and the road surface slope angle λ(negatively shown) according to ISO 8855:2011. F_(X) _(R,cg), F_(Y)_(R,cg) and F_(Z) _(R,cg) are arbitrary force vector components, whichare aligned with the road surface axle system and operate at the centerof gravity of the vehicle 1. M_(X) _(R,cg), M_(Y) _(R,cg) and M_(Z)_(R,cg) are arbitrary torque vector components, which are aligned withthe road surface axle system.

FIG. 2 further shows a roll angle φ (positively shown, rotation aroundthe X_(V) axis), the vehicle roll angle φ_(V) (positively shown,rotation around the X axis), and the road surface-level camber angle η(positively shown, rotation around the X axis), according to ISO8855:2011.

FIG. 3 further shows the yaw angle _(Ψ) (positively shown, from theX_(E) axis to the X axis, around Z_(E)) and the left and right frontsteering angle δ_(1L), δ_(1R) (positively shown, angle from the X_(V)axis to the wheel plane, around the Z_(V) axis), according to ISO8855:2011.

Acceleration and speed of potential trajectories that are providedthrough the trajectory planning are given as projections of the desiredmovement of the rear axle in the interconnected X-Y plane. So that theycan be used as input parameters of a model, these characteristics mustbe converted to accelerations and speeds describing the movement of thecenter of gravity in the road surface plane given by X_(R)-Y_(R). Thecorresponding coordinate transformation between the coordinate systems(x, y, z) and (x_(R), y_(R), z_(R)) are derived in this phase.

FIG. 1 shows the corresponding rotations. It should be noted that theangles provided as input parameters are the road surface-level slope λ(gradients) and road surface-level camber angle η (transverse incline),which are not equal to the rotation angles that are needed for the axlerotations. The following features can be taken from FIG. 1 :

$\begin{matrix}{\overline{OB} = 1,} & \text{­­­(1)}\end{matrix}$

$\begin{matrix}{\overline{OC} = 1,} & \text{­­­(2)}\end{matrix}$

$\begin{matrix}{\overline{AB} = \sin\left( \eta_{XR} \right),} & \text{­­­(3)}\end{matrix}$

$\begin{matrix}{\overline{DC} = \sin(\eta),} & \text{­­­(4)}\end{matrix}$

$\begin{matrix}{\overline{AC} = \overline{AB} = \sin\left( \eta_{XR} \right),} & \text{­­­(5)}\end{matrix}$

$\begin{matrix}{\frac{\overline{DC}}{\overline{AC}} = \cos(\lambda),} & \text{­­­(6)}\end{matrix}$

$\begin{matrix}{\overline{OE} = \cos\left( \eta_{XR} \right),} & \text{­­­(7)}\end{matrix}$

$\begin{matrix}{\overline{EF} = \sin\left( \eta_{XR} \right),} & \text{­­­(8)}\end{matrix}$

$\begin{matrix}{\overline{TF} = \overline{OE} = \cos\left( \eta_{XR} \right),} & \text{­­­(9)}\end{matrix}$

$\begin{matrix}{\overline{IG} = \overline{IF} = \cos\left( \eta_{XR} \right).} & \text{­­­(10)}\end{matrix}$

From this follows:

$\begin{matrix}{\cos(\lambda) = \frac{\overline{DC}}{\overline{AC}} = \frac{\sin(\eta)}{\sin\left( \eta_{XR} \right)},} & \text{­­­(11)}\end{matrix}$

$\begin{matrix}{\sin\left( \eta_{XR} \right) = \frac{\sin(\eta)}{\cos(\lambda)},} & \text{­­­(12)}\end{matrix}$

$\begin{matrix}{\sin(\eta) = \sin\left( \eta_{XR} \right) \cdot \cos(\lambda).} & \text{­­­(13)}\end{matrix}$

FIG. 4 shows the rotations of the axles that put the road surface-levelslope angle λ and the road surface-level camber angle η in relation:

1. The rotation of the interconnected, levelled axle system (X, Y, Z) atthe angle ηX R around X leads to an auxiliary axle system (X′, Y′, Z′).

2. The rotation of the auxiliary axle system (X′, Y′, Z′) around Y (!)around the angle λ results in the road surface axle system (X_(R),Y_(R), Z_(R)). It was noticed that the rotation occurs around theoriginal Y axis, not around Y′, which is different to the usual Eulerangle or Tait-Bryan angle rotations, which cause rotations around theresulting axle systems. λ and η are the angles between X_(R), Y_(R) andtheir respective projections in the horizontal plane. These projectionsdo not form a perpendicular pair of vectors.

In order to define coordinate transformations (or passivetransformations, as in M.J. Benacquista, J.D. Romano, ClassicalMechanics, Springer, Cham, CH, 2018, pp. 193) between (x, y, z) and(X_(R), Y_(R), Z_(R)), it is possible to differentiate between intrinsicand extrinsic rotation matrices (see M.J. Benacquista, J.D. Romano,Classical Mechanics, Springer, Cham, CH, 2018, pp. 200), which aredefined either with axes relative to the rotated body (intrinsically) orwith axes that are fixed in space (extrinsically). The rotations in FIG.4 are given as a sequence of extrinsic rotations:

1. Rotation around X, around the angle η_(X) _(R).

2. Rotation around Y, around the angle λ.

Applying M.J. Benacquista, J.D. Romano, Classical Mechanics, Springer,Cham, CH, 2018, this is equivalent to the intrinsic rotations:

-   1. Rotation around Y, around the angle λ,-   2. Rotation around X₀ (!), around the angle η_(X) _(R).

Therefore, the coordinate transformation T_(IR) from (x, y, z) to(x_(R), y_(R), z_(R)) can be given using two rotation matrices forintrinsic rotations, for example:

$\begin{matrix}{\begin{bmatrix}x_{R} \\y_{R} \\z_{R}\end{bmatrix} = \underset{\text{T}_{IR}}{\underset{︸}{\underset{\text{T}_{X},{(\eta_{XR})}}{\underset{︸}{\left\lbrack \begin{array}{lll}1 & 0 & 0 \\0 & {\cos\left( \eta_{XR} \right)} & {\sin\left( \eta_{XR} \right)} \\0 & {- \sin\left( \eta_{XR} \right)} & {\cos\left( \eta_{XR} \right)}\end{array} \right\rbrack}}\underset{T_{Y{(\lambda)}}}{\underset{︸}{\left\lbrack \begin{array}{lll}{\cos(\lambda)} & 0 & {- \sin(\lambda)} \\0 & 1 & 0 \\{\sin(\lambda)} & 0 & {\cos(\lambda)}\end{array} \right\rbrack}}}}\begin{bmatrix}x \\y \\z\end{bmatrix}} & \text{­­­(14)}\end{matrix}$

By combining both transformation matrices, T_(IR) is given by:

$\begin{matrix}{\text{T}_{IR} = \left\lbrack \begin{array}{lll}{\cos(\lambda)} & 0 & {- \sin(\lambda)} \\{\sin\left( \eta_{XR} \right)\sin(\lambda)} & {\cos\left( \eta_{XR} \right)} & {\sin\left( \eta_{XR} \right)\cos(\lambda)} \\{\cos\left( \eta_{XR} \right)\sin(\lambda)} & {- \sin\left( \eta_{XR} \right)} & {\cos\left( \eta_{XR} \right)\cos(\lambda)}\end{array} \right\rbrack} & \text{­­­(15)}\end{matrix}$

Replacing η_(XR) with equation (12) gives:

$\begin{matrix}\begin{array}{l}{\text{T}_{IR} =} \\\left\lbrack \begin{array}{lll}{\cos(\lambda)} & 0 & {- \sin(\lambda)} \\{\frac{\sin(\eta)}{\cos(\lambda)}\sin(\lambda)} & {\cos\left( {\arcsin\left( \frac{\sin(\eta)}{\cos(\lambda)} \right)} \right)} & {\sin(\eta)} \\{\cos\left( {\arcsin\left( \frac{\sin(\eta)}{\cos(\lambda)} \right)} \right)\sin(\lambda)} & {- \frac{\sin(\eta)}{\cos(\lambda)}} & {\cos\left( {\arcsin\left( \frac{\sin(\eta)}{\cos(\lambda)} \right)} \right)\cos(\lambda)}\end{array} \right\rbrack\end{array} & \text{­­­(16)}\end{matrix}$

The coordinate transformation T_(ER) of fixed coordinates (x_(E), y_(E),z_(E)) to road surface-level coordinates (x_(R), y_(R), z_(R)) can befound by extending equation (14):

$\begin{matrix}\begin{array}{l}{\left\lbrack \begin{array}{l}x_{R} \\y_{R} \\z_{R}\end{array} \right\rbrack =} \\{\underset{\text{T}_{\Xi R}}{\underset{︸}{\underset{\text{T}_{X},{(\eta_{XR})}}{\underset{︸}{\left\lbrack \begin{array}{lll}1 & 0 & 0 \\0 & {\cos\left( \eta_{XR} \right)} & {\sin\left( \eta_{XR} \right)} \\0 & {- \sin\left( \eta_{XR} \right)} & {\cos\left( \eta_{XR} \right)}\end{array} \right\rbrack}}\underset{T_{Y{(\lambda)}}}{\underset{︸}{\left\lbrack \begin{array}{lll}{\cos(\lambda)} & 0 & {- \sin(\lambda)} \\0 & 1 & 0 \\{\sin(\lambda)} & 0 & {\cos(\lambda)}\end{array} \right\rbrack}}\underset{Tz_{\Xi}{(\psi)}}{\underset{︸}{\left\lbrack \begin{array}{lll}{\cos(\psi)} & {\sin(\psi)} & 0 \\{- \sin(\psi)} & {\cos(\psi)} & 0 \\0 & 0 & 1\end{array} \right\rbrack}}}}\left\lbrack \begin{array}{l}x_{\Xi} \\y_{\Xi} \\z_{\Xi}\end{array} \right\rbrack}\end{array} & \text{­­­(17)}\end{matrix}$

T_(ZE) thereby refers to the matrix in right rotation around Z_(E),around the angle _(Ψ) (see FIG. 3 ) connected with a coordinatetransformation of fixed coordinates (x_(E), y_(E), z_(E)) tointermediate coordinates (x, y, z), as can be found in D.T. Greenwood,Principles of Dynamics, Prentice Hall, Upper Saddle River, US-NJ, 2ndedition, 1988, pp. 357. With the combination of all transformationmatrices, T_(ER) is given by:

$\begin{matrix}\begin{array}{l}{\text{T}_{ER} = \left\lbrack \begin{array}{l}{\cos(\psi)\cos(\lambda)} \\{- \sin(\psi)\cos\left( \eta_{XR} \right) + \cos(\psi)\sin(\lambda)\sin\left( \eta_{XR} \right)} \\{\sin(\psi)\sin\left( \eta_{XR} \right) + \cos(\psi)\sin(\lambda)\cos\left( \eta_{XR} \right)}\end{array} \right)} \\\left( \begin{array}{ll}{\sin\left( \text{ψ} \right)\cos\left( \text{λ} \right)} & {- \sin\left( \text{λ} \right)} \\{\cos\left( \text{ψ} \right)\cos\left( \text{η}_{\text{XR}} \right) + \sin\left( \text{ψ} \right)\sin\left( \text{λ} \right)\sin\left( \text{η}_{\text{XR}} \right)} & {\cos\left( \text{λ} \right)\sin\left( \text{η}_{\text{XR}} \right)} \\{- \cos\left( \text{ψ} \right)\sin\left( \text{η}_{\text{XR}} \right) + \sin\left( \text{ψ} \right)\sin\left( \text{λ} \right)\cos\left( \text{η}_{\text{XR}} \right)} & {\cos\left( \text{λ} \right)\cos\left( \text{η}_{\text{XR}} \right)}\end{array} \right\rbrack\end{array} & \text{­­­(18)}\end{matrix}$

The matrix in equation (18) contains a line break to improvereadability, wherein the first column before the line break and thesecond and third column after the line break are shown.

For the quasi-steady-state descriptions of the vehicle, a double trackmodel in motion is assumed as an assumed model.

The following aspects are taken into consideration:

-   lateral acceleration in curves-   longitudinal acceleration as a result of driving and braking-   aerodynamic drag-   aerodynamic lift-   roll torque distribution between front axle and rear axle-   stationary influence of transverse incline and gradient-   roll resistance (only the influence of the necessary motor torque)

The following aspects are disregarded:

-   roll and pitching movement of the suspension-   fast changes of transverse incline and gradient-   differences between overall mass and sprung mass-   asymmetric lateral load conditions-   influence of ABS, ASR and ESP (i.e., the model is a slightly    conservative description of the vehicle capabilities)-   differences in left and right driving and braking forces (i.e. no    torque vectoring and no p-split)-   long cornering and parking maneuvers

FIG. 5 is a schematic view of the vehicle 1 in the X_(R)-Z_(R) plane,wherein the roll axis RA, the front roll center FRC and the rear rollcenter RRC are illustrated. h_(cg) refers to the height of the center ofgravity, I_(f) the space between the center of gravity and the frontaxle and I the wheelbase. F_(X R,f), F_(Z R,f), F_(X R,r), F_(Z) _(R,r)are axle forces and F_(X) _(R,R1), F_(Z) _(R,R1), F_(X R,L1),F_(Z R,L1), F_(X) _(R,R2), F_(Z) _(R,R2), F_(X) _(R,L2), F_(Z R,L2) aretire forces. F_(X) _(R,cg) and F_(Z) _(R,cg) are arbitrary force vectorcomponents that are aligned with the road surface plane axle system atthe center of gravity of the vehicle 1. M_(Y) _(R,cg) is an arbitrarytorque vector component that is aligned with the road surface plane axlesystem.

It is apparent from FIG. 5 that the balance of forces in the ZRdirection result as follows:

$\begin{matrix}{\text{F}_{\text{Z}\mspace{6mu}\text{R,f}} + \text{F}_{\text{Z}\mspace{6mu}\text{R,r}} + \text{F}_{\text{Z}\mspace{6mu}\text{R,cg}} = 0.} & \text{­­­(19)}\end{matrix}$

The balance of the torques around Y_(R) on the front axle at groundlevel is:

$\begin{matrix}{\text{F}_{\text{Z}\mspace{6mu}\text{R,r}} \cdot \text{I} + \text{F}_{\text{Z}\mspace{6mu}\text{R,cg}} \cdot \text{If} + \text{F}_{\text{X}\mspace{6mu}\text{R,cg}} \cdot \text{hcg}\mspace{6mu}\text{+M}_{\text{Y}\mspace{6mu}\text{R,cg}} = 0.} & \text{­­­(20)}\end{matrix}$

From (20) follows:

$\begin{matrix}{F_{ZR,r} = - \frac{l_{f}}{l}F_{ZR,cg} - \frac{h_{cg}}{l}F_{XR,cg} - \frac{1}{l}M_{YR,cg}} & \text{­­­(21)}\end{matrix}$

and application in equation (19) leads to:

$\begin{matrix}{F_{ZR,f} = - \underset{= \frac{l_{h}}{l}}{\underset{︸}{\left( {1 - \frac{l_{f}}{l}} \right)}}F_{ZR,cg} + \frac{h_{cg}}{l}F_{XR,cg} + \frac{1}{l}M_{YR,cg}.} & \text{­­­(22)}\end{matrix}$

The plausibility can easily be considered by applying, for example,

F_(Z R,cg) = −m ⋅ g and F_(X R,cg) = −m ⋅ a_(X R).

Note: Wheel load transfer during acceleration and braking is assumed ina uniformly distributed manner on the left and right wheel due tosymmetry. Therefore, in (21) and (22), no pitching center or pitchingtorque distribution must be taken into consideration. However, the rolltorque distribution is taken into consideration in the following phases.

FIG. 6 shows a schematic view of the vehicle 1 in the X_(R)-Y_(R) plane.F_(X R,f), F_(Y) _(R,f), F_(X R,r), F_(Y) _(R,r) are axle forces. F_(X)_(R,cg), F_(Y) _(R,cg) and F_(Z) _(R,cg) are arbitrary force vectorcomponents, which are aligned with the road surface plane axle systemand operate at the center of gravity of the vehicle 1. M_(X) _(R,cg),M_(Y) _(R,cg) and M_(Z) _(R,cg) are arbitrary torque vector components,which are aligned with the road surface plane axle system.

It is apparent from FIG. 6 that the balance of forces in the YRdirection result as follows:

$\begin{matrix}{\text{F}_{\text{Y}\mspace{6mu}\text{R,f}} + \text{F}_{\text{Y}\mspace{6mu}\text{R,r}} + \text{F}_{\text{Y}\mspace{6mu}\text{R,cg}} = 0.} & \text{­­­(23)}\end{matrix}$

The balance of the torques around Z_(R) on the front axle center is:

$\begin{matrix}{- \text{F}_{\text{Y}\mspace{6mu}\text{R,cg}} \cdot \text{I}_{\text{f}} - \text{F}_{\text{Y}\mspace{6mu}\text{R,r}} \cdot \text{I} + \text{M}_{\text{Z}\mspace{6mu}\text{R,cg}} = 0.} & \text{­­­(24)}\end{matrix}$

Left and right braking and driving forces are assumed equally andtherefore do not contribute to the balance of the torques around Z_(R).It therefore follows from (24) that

$\begin{matrix}{F_{YR,r} = - \frac{l_{f}}{l}F_{YR,cg} + \frac{1}{l}M_{ZR,cg}} & \text{­­­(25)}\end{matrix}$

and applying this equation in (25) leads to

$\begin{matrix}{F_{YR,f} = - \underset{= \frac{l_{h}}{l}}{\underset{︸}{\left( {1 - \frac{l_{f}}{l}} \right)}}F_{YR,cg} - \frac{1}{l}M_{ZR,cg}.} & \text{­­­(26)}\end{matrix}$

In the X_(R) direction, this applies:

$\begin{matrix}{\text{F}_{\text{X}\mspace{6mu}\text{R,f}} + \text{F}_{\text{X}\mspace{6mu}\text{R,r}} + \text{F}_{\text{X}\mspace{6mu}\text{R,cg}} = 0.} & \text{­­­(27)}\end{matrix}$

F_(X R,r) and F_(X R,f) are determined by driving and braking torquedistribution.

FIG. 7 is a schematic view of the vehicle 1 in the Y_(R)-Z_(R) planewith the front and rear axle forces F_(Y) _(R,f), F_(Y) _(R,r) andtorques M_(X) _(R,f), M_(X R,r). F_(Y) _(R,cg), and F_(Z R,cg) arearbitrary force vector components, which are aligned with the roadsurface plane axle system and operate at the center of gravity of thevehicle 1. M_(X) _(R,cg) is an arbitrary torque vector component, whichis aligned with the road surface plane axle system. h_(cg) refers to theheight of the center of gravity and h_(rcg) is the height of the rollcenter RC_(cg) at the x_(R) position of the center of gravity.

The balance of the torques around the X_(R) axis can be taken from FIG.7 , which illustrates the front and rear axle forces F_(Y) _(R,f), F_(Y)_(R,r) and torques M_(X) _(R,f), M_(X R,r) as well as arbitrary forcesand torques that operate at the center of gravity. If the roll centerRC_(cg) is used at the X_(R) position of the center of gravity in orderto convey the balance of the torques around x_(R), then it follows that:

$\begin{matrix}{M_{XR,cg} + \underset{: = M_{XR,f + r}}{\underset{︸}{M_{XR,f} + M_{XR,r}}} - \left( {h_{cg} - h_{rcg}} \right) \cdot F_{YR,cg} = 0.} & \text{­­­(28)}\end{matrix}$

With the introduction of M_(X) _(R,f+r), this can be transposed as:

$\begin{matrix}{\text{M}_{\text{X}\mspace{6mu}\text{R,f+r}} = \left( {\text{h}_{\text{cg}} - \text{h}_{\text{rcg}}} \right) \cdot \text{F}_{\text{Y}\mspace{6mu}\text{R,cg}} - \text{M}_{\text{X}\mspace{6mu}\text{R,cg}}.} & \text{­­­(29)}\end{matrix}$

FIG. 8 is a schematic view of the front axle in the Y_(R)-Z_(R) planewith the tire forces. F_(Y) _(R,1L), F_(Z) _(R,1L), F_(YR,1R),F_(Z R,1R), chassis reaction forces F_(Y) _(R,1), F_(Z R,1) and thechassis reaction torque M_(X) _(R,1). h_(rf) refers to the height of theroll center of the front axle FRC and b_(f) is the track width of thefront axle. It should be noted that the tire forces F_(Y) _(R,1L) andF_(Y) _(a,1a) are still given as vector components in road surface planecoordinates, which do not correspond to F_(Y T,1L) and F_(Y T,1R) intire coordinates.

If a roll torque distribution is introduced that combines the influenceof the stiffness of the suspension and of the stabilizer, it is possibleto maintain the roll torques of the front and rear axles that are causeddue to the displacement between the center of gravity and the roll axis:

$\begin{matrix}{M_{XR,f} = \kappa \cdot M_{XR,f + r} = \kappa\left( {\left( {h_{cg} - h_{rcg}} \right) \cdot F_{YR,cg} - M_{XR,cg}} \right),} & \text{­­­(30)}\end{matrix}$

$\begin{matrix}{M_{XR,r} = \left( {1 - \kappa} \right) \cdot M_{XR,f + r} = \left( {1 - \kappa} \right)\left( {\left( {h_{cg} - h_{rcg}} \right) \cdot F_{YR,cg} - M_{XR,cg}} \right).} & \text{­­­(31)}\end{matrix}$

Since the roll torque distribution and the axle forces are provided, itis possible to calculate the vertical tire forces as illustrated in FIG.8 .

The balance of forces in the Z_(R) and Y_(R) directions results in:

$\begin{matrix}{\text{F}_{\text{Z}\mspace{6mu}\text{R,1L}} + \text{F}_{\text{Z}\mspace{6mu}\text{R,1R}} + \text{F}_{\text{Z}\mspace{6mu}\text{R,1}} = 0.} & \text{­­­(32)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{Y}\mspace{6mu}\text{R,1L}} + \text{F}_{\text{Y}\mspace{6mu}\text{R,1R}} + \text{F}_{\text{Y}\mspace{6mu}\text{R,1}} = 0.} & \text{­­­(33)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{Z}\mspace{6mu}\text{R,2L}} + \text{F}_{\text{Z}\mspace{6mu}\text{R,2R}} + \text{F}_{\text{Z}\mspace{6mu}\text{R,2}} = 0.} & \text{­­­(34)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,2L}} + \text{F}_{\text{YR,2R}} + \text{F}_{\text{YR,2}} = 0.} & \text{­­­(35)}\end{matrix}$

The balance of torques around the axis of symmetry at the road surfaceplane level results in:

$\begin{matrix}{\frac{b_{f}}{2}F_{ZR,1L} - \frac{b_{f}}{2}F_{ZR,1R} - h_{rf} \cdot F_{YR,1} + M_{XR,1} = 0} & \text{­­­(36)}\end{matrix}$

$\begin{matrix}{\frac{b_{r}}{2}F_{ZR,2L} - \frac{b_{r}}{2}F_{ZR,2R} - h_{rr} \cdot F_{YR,2} + M_{XR,2} = 0.} & \text{­­­(37)}\end{matrix}$

With Newton’s third law, the chassis reaction forces and the axle forcesare put in relation as follows:

$\begin{matrix}{\text{F}_{\text{YR,1}} = - \text{F}_{\text{YR,f}},} & \text{­­­(38)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{ZR,1}} = - \text{F}_{\text{ZR,f}},} & \text{­­­(39)}\end{matrix}$

$\begin{matrix}{\text{M}_{\text{XR,1}} = - \text{M}_{\text{XR,f}},} & \text{­­­(40)}\end{matrix}$

The application of this and (32) to (37) leads to:

$\begin{matrix}{\text{F}_{\text{ZR,1L}} + \text{F}_{\text{ZR,1R}} - \text{F}_{\text{ZR,f}} = 0,} & \text{­­­(41)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,1L}} + \text{F}_{\text{YR,1R}} - \text{F}_{\text{YR,f}} = 0,} & \text{­­­(42)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{ZR,2L}} + \text{F}_{\text{ZR,2R}} - \text{F}_{\text{ZR,r}} = 0,} & \text{­­­(43)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,2L}} + \text{F}_{\text{YR,2R}} - \text{F}_{\text{YR,r}} = 0,} & \text{­­­(44)}\end{matrix}$

$\begin{matrix}{\frac{\text{b}_{\text{f}}}{2}\underset{: = 2 \cdot \text{Δ}\text{F}_{\text{ZR,1}}}{\underset{︸}{\left( {\text{F}_{\text{ZR,1L}} - \text{F}_{\text{ZR,1R}}} \right)}} + \text{h}_{\text{rf}} \cdot \text{F}_{\text{YR,f}} - \text{M}_{\text{XR,f}} = 0,} & \text{­­­(45)}\end{matrix}$

$\begin{matrix}{\frac{\text{b}_{\text{r}}}{2}\underset{: = 2 \cdot \text{Δ}\text{F}_{\text{ZR,2}}}{\underset{︸}{\left( {\text{F}_{\text{ZR,2L}} - \text{F}_{\text{ZR,2R}}} \right)}} + \text{h}_{\text{rr}} \cdot \text{F}_{\text{YR,r}} - \text{M}_{\text{XR,r}} = 0.} & \text{­­­(46)}\end{matrix}$

The solution of (41) and (43) as well as for F_(Z) _(R,1 L), F_(Z R,1R)and also for F_(Z) _(R,2L), F_(Z R,2R) and the subsequent application ofthe results in (45) and (46) results in:

$\begin{matrix}{\frac{\text{b}_{\text{f}}}{2}\left( {\text{F}_{\text{ZR,1L}} - \text{F}_{\text{ZR,f}} + \text{F}_{\text{ZR,1L}}} \right) + \text{h}_{\text{rf}} \cdot \text{F}_{\text{YR,f}} - \text{M}_{\text{XR,f}} = 0,} & \text{­­­(47)}\end{matrix}$

$\begin{matrix}{\text{b}_{\text{f}} \cdot \text{F}_{\text{ZR,1L}} - \frac{\text{b}_{\text{f}}}{2}\text{F}_{\text{ZR,f}} + \text{h}_{\text{rf}} \cdot \text{F}_{\text{YR,f}} - \text{M}_{\text{XR,f}} = 0,} & \text{­­­(48)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{ZR,1L}} = \frac{1}{2}\text{F}_{\text{ZR,f}} - \frac{\text{h}_{\text{rf}}}{\text{b}_{\text{f}}}\text{F}_{\text{YR,f}} + \frac{1}{\text{b}_{\text{f}}}\text{M}_{\text{XR,f}}} & \text{­­­(49)}\end{matrix}$

$\begin{matrix}{\frac{\text{b}_{\text{f}}}{2}\left( {\text{F}_{\text{ZR,f}} - \text{F}_{\text{ZR,1R}} - \text{F}_{\text{ZR,1R}}} \right) + \text{h}_{\text{rf}} \cdot \text{F}_{\text{YR,f}} - \text{M}_{\text{XR,f}} = 0,} & \text{­­­(50)}\end{matrix}$

$\begin{matrix}{\frac{\text{b}_{\text{f}}}{2}\text{F}_{\text{ZR,f}} - \text{b}_{\text{f}} \cdot \text{F}_{\text{ZR,1R}} + \text{h}_{\text{rf}} \cdot \text{F}_{\text{YR,f}} - \text{M}_{\text{XR,f}} = 0,} & \text{­­­(51)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{ZR,1R}} = \frac{1}{2}\text{F}_{\text{ZR,f}} + \frac{\text{h}_{\text{rf}}}{\text{b}_{\text{f}}}\text{F}_{\text{YR,f}} - \frac{1}{\text{b}_{\text{f}}}\text{M}_{\text{XR,f}}} & \text{­­­(52)}\end{matrix}$

$\begin{matrix}{\frac{\text{b}_{\text{r}}}{2}\left( {\text{F}_{\text{ZR,2L}} - \text{F}_{\text{ZR,r}} + \text{F}_{\text{ZR,2L}}} \right) + \text{h}_{\text{rr}} \cdot \text{F}_{\text{YR,r}} - \text{M}_{\text{XR,r}} = 0,} & \text{­­­(53)}\end{matrix}$

$\begin{matrix}{\text{b}_{\text{r}} \cdot \text{F}_{\text{ZR,2L}} - \frac{\text{b}_{\text{r}}}{2}\text{F}_{\text{ZR,r}} + \text{h}_{\text{rr}} \cdot \text{F}_{\text{YR,r}} - \text{M}_{\text{XR,r}} = 0,} & \text{­­­(54)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{ZR,2L}} = \frac{1}{2}\text{F}_{\text{ZR,r}} - \frac{\text{h}_{\text{rr}}}{\text{b}_{\text{r}}}\text{F}_{\text{YR,r}} + \frac{1}{\text{b}_{\text{r}}}\text{M}_{\text{XR,r}}} & \text{­­­(55)}\end{matrix}$

$\begin{matrix}{\frac{\text{b}_{\text{r}}}{2}\left( {\text{F}_{\text{ZR,r}} - \text{F}_{\text{ZR,2R}} - \text{F}_{\text{ZR,2R}}} \right) + \text{h}_{\text{rr}} \cdot \text{F}_{\text{YR,r}} - \text{M}_{\text{XR,r}} = 0,} & \text{­­­(56)}\end{matrix}$

$\begin{matrix}{\frac{\text{b}_{\text{r}}}{2}\text{F}_{\text{ZR,r}} - \text{b}_{\text{r}} \cdot \text{F}_{\text{ZR,2R}} + \text{h}_{\text{rr}} \cdot \text{F}_{\text{YR,r}} - \text{M}_{\text{XR,r}} = 0,} & \text{­­­(57)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{ZR,2R}} = \frac{1}{2}\text{F}_{\text{ZR,r}} + \frac{\text{h}_{\text{rr}}}{\text{b}_{\text{r}}}\text{F}_{\text{YR,r}} - \frac{1}{\text{b}_{\text{r}}}\text{M}_{\text{XR,r}}} & \text{­­­(58)}\end{matrix}$

Application of (22), (26) and (30) in (49) and (52) results in:

$\begin{matrix}\begin{array}{l}{\text{F}_{\text{ZR,1L}} = \frac{1}{2}\left( {- \left( {1 - \frac{\text{l}_{\text{f}}}{\text{l}}} \right)\text{F}_{\text{ZR,cg}} + \frac{\text{h}_{\text{cg}}}{\text{l}}\text{F}_{\text{XR,cg}} + \frac{1}{\text{l}}\text{M}_{\text{YR,cg}}} \right)} \\{- \frac{\text{h}_{\text{rf}}}{\text{b}_{\text{f}}}\left( {- \left( {1 - \frac{\text{l}_{\text{f}}}{\text{l}}} \right)\text{F}_{\text{YR,cg}} - \frac{1}{\text{l}}\text{M}_{\text{ZR,cg}}} \right)} \\{+ \frac{1}{\text{b}_{\text{f}}} \cdot \text{κ}\left( {\left( {\text{h}_{\text{cg}} - \text{h}_{\text{rcg}}} \right)\text{F}_{\text{YR,cg}} - \text{M}_{\text{XR,cg}}} \right),}\end{array} & \text{­­­(59)}\end{matrix}$

$\begin{matrix}\begin{array}{l}{\underset{\geq 0{(\text{constraint})}}{\underset{︸}{\text{F}_{\text{ZR,1L}}}} = \frac{1}{2}\frac{\text{h}_{\text{cg}}}{\text{l}}\text{F}_{\text{XR,cg}} - \frac{1}{\text{b}_{\text{f}}} \cdot \text{κ} \cdot \text{M}_{\text{XR,cg}}} \\{+ \left( {\frac{\text{h}_{\text{rf}}}{\text{b}_{\text{f}}}\left( {1 - \frac{\text{l}_{\text{f}}}{\text{l}}} \right) + \text{κ}\frac{\text{h}_{\text{cg}} - \text{h}_{\text{rcg}}}{\text{b}_{\text{f}}}} \right)\text{F}_{\text{YR,cg}} + \frac{1}{2 \cdot \text{l}}\text{M}_{\text{YR,cg}}} \\{- \frac{1}{2}\left( {1 - \frac{\text{l}_{\text{f}}}{\text{l}}} \right)\text{F}_{\text{ZR,cg}} + \frac{\text{h}_{\text{rf}}}{\text{b}_{\text{f}} \cdot 1}\text{M}_{\text{ZR,cg}},}\end{array} & \text{­­­(60)}\end{matrix}$

$\begin{matrix}\begin{array}{l}{\text{F}_{\text{ZR,1R}} = \frac{1}{2}\left( {- \left( {1 - \frac{\text{l}_{\text{f}}}{\text{l}}} \right)\text{F}_{\text{ZR,cg}} + \frac{\text{h}_{\text{cg}}}{\text{l}}\text{F}_{\text{XR,cg}} + \frac{1}{\text{l}}\text{M}_{\text{YR,cg}}} \right)} \\{+ \frac{\text{h}_{\text{rf}}}{\text{b}_{\text{f}}}\left( {- \left( {1 - \frac{\text{l}_{\text{f}}}{\text{l}}} \right)\text{F}_{\text{YR,cg}} - \frac{1}{\text{l}}\text{M}_{\text{ZR,cg}}} \right)} \\{- \frac{1}{\text{b}_{\text{f}}} \cdot \text{κ}\left( {\left( {\text{h}_{\text{cg}} - \text{h}_{\text{rcg}}} \right)\text{F}_{\text{YR,cg}} - \text{M}_{\text{XR,cg}}} \right),}\end{array} & \text{­­­(61)}\end{matrix}$

$\begin{matrix}\begin{array}{l}{\underset{\geq 0{(\text{constraint})}}{\underset{︸}{\text{F}_{\text{ZR,1R}}}} = \frac{1}{2}\frac{\text{h}_{\text{cg}}}{\text{l}}\text{F}_{\text{XR,cg}} + \frac{1}{\text{b}_{\text{f}}} \cdot \text{κ} \cdot \text{M}_{\text{XR,cg}}} \\{- \left( {\frac{\text{h}_{\text{rf}}}{\text{b}_{\text{f}}}\left( {1 - \frac{\text{l}_{\text{f}}}{\text{l}}} \right) + \text{κ}\frac{\text{h}_{\text{cg}} - \text{h}_{\text{rcg}}}{\text{b}_{\text{f}}}} \right)\text{F}_{\text{YR,cg}} + \frac{1}{2 \cdot \text{l}}\text{M}_{\text{YR,cg}}} \\{- \frac{1}{2}\left( {1 - \frac{\text{l}_{\text{f}}}{\text{l}}} \right)\text{F}_{\text{ZR,cg}} - \frac{\text{h}_{\text{rf}}}{\text{b}_{\text{f}} \cdot \text{l}}\text{M}_{\text{ZR,cg}},}\end{array} & \text{­­­(62)}\end{matrix}$

Application of (21), (25) and (31) in (55) and (58) results in:

$\begin{matrix}\begin{array}{l}{\text{F}_{\text{ZR,2L}} = \frac{1}{2}\left( {- \frac{\text{I}_{\text{f}}}{\text{l}}\text{F}_{\text{ZR,cg}} - \frac{\text{h}_{\text{cg}}}{\text{l}}\text{F}_{\text{XR,cg}} - \frac{1}{\text{l}}\text{M}_{\text{YR,cg}}} \right)} \\{- \frac{\text{h}_{\text{rr}}}{\text{b}_{\text{r}}}\left( {- \frac{\text{l}_{\text{f}}}{\text{l}}\text{F}_{\text{YR,cg}} + \frac{1}{\text{l}}\text{M}_{\text{ZR,cg}}} \right)} \\{+ \frac{1}{\text{b}_{\text{r}}} \cdot \left( {1 - \text{κ}} \right)\left( {\left( {\text{h}_{\text{cg}} - \text{h}_{\text{rcg}}} \right)\text{F}_{\text{YR,cg}} - \text{M}_{\text{XR,cg}}} \right),}\end{array} & \text{­­­(63)}\end{matrix}$

$\begin{matrix}\begin{array}{l}{\underset{\geq 0{(\text{constraint})}}{\underset{︸}{\text{F}_{\text{ZR,2L}}}} = - \frac{1}{2}\frac{\text{h}_{\text{cg}}}{\text{l}}\text{F}_{\text{XR,cg}} - \frac{1}{\text{b}_{\text{r}}} \cdot \left( {1 - \text{κ}} \right) \cdot \text{M}_{\text{XR,cg}}} \\{+ \left( {\frac{\text{h}_{\text{rr}}}{\text{b}_{\text{r}}}\frac{\text{l}_{\text{f}}}{\text{l}} + \left( {1 - \text{κ}} \right)\frac{\text{h}_{\text{cg}} - \text{h}_{\text{rcg}}}{\text{b}_{\text{r}}}} \right)\text{F}_{\text{YR,cg}} - \frac{1}{2 \cdot \text{l}}\text{M}_{\text{YR,cg}}} \\{- \frac{1}{2}\frac{\text{l}_{\text{f}}}{\text{l}}\text{F}_{\text{ZR,cg}} - \frac{\text{h}_{\text{rr}}}{\text{b}_{\text{r}} \cdot 1}\text{M}_{\text{ZR,cg}},}\end{array} & \text{­­­(64)}\end{matrix}$

$\begin{matrix}\begin{array}{l}{\text{F}_{\text{ZR,2R}} = \frac{1}{2}\left( {- \frac{\text{l}_{\text{f}}}{\text{l}}\text{F}_{\text{ZR,cg}} - \frac{\text{h}_{\text{cg}}}{\text{l}}\text{F}_{\text{XR,cg}} - \frac{1}{\text{l}}\text{M}_{\text{YR,cg}}} \right)} \\{+ \frac{\text{h}_{\text{rr}}}{\text{b}_{\text{r}}}\left( {- \frac{\text{l}_{\text{f}}}{\text{l}}\text{F}_{\text{YR,cg}} + \frac{1}{\text{l}}\text{M}_{\text{ZR,cg}}} \right)} \\{+ \frac{1}{\text{b}_{\text{r}}} \cdot \left( {1 - \text{κ}} \right)\left( {\left( {\text{h}_{\text{cg}} - \text{h}_{\text{rcg}}} \right)\text{F}_{\text{YR,cg}} - \text{M}_{\text{XR,cg}}} \right),}\end{array} & \text{­­­(65)}\end{matrix}$

$\begin{matrix}\begin{array}{l}{\underset{\geq 0{(\text{constraint})}}{\underset{︸}{\text{F}_{\text{ZR,2R}}}} = - \frac{1}{2}\frac{\text{h}_{\text{cg}}}{\text{l}}\text{F}_{\text{XR,cg}} + \frac{1}{\text{b}_{\text{r}}} \cdot \left( {1 - \text{κ}} \right) \cdot \text{M}_{\text{XR,cg}}} \\{+ \left( {\frac{\text{h}_{\text{rr}}}{\text{b}_{\text{r}}}\frac{\text{l}_{\text{f}}}{\text{l}} + \left( {1 - \text{κ}} \right)\frac{\text{h}_{\text{cg}} - \text{h}_{\text{rcg}}}{\text{b}_{\text{r}}}} \right)\text{F}_{\text{YR,cg}} - \frac{1}{2 \cdot \text{l}}\text{M}_{\text{YR,cg}}} \\{- \frac{1}{2}\frac{\text{l}_{\text{f}}}{\text{l}}\text{F}_{\text{ZR,cg}} + \frac{\text{h}_{\text{rr}}}{\text{b}_{\text{r}} \cdot 1}\text{M}_{\text{ZR,cg}}.}\end{array} & \text{­­­(66)}\end{matrix}$

F_(Z) _(R,1L), F_(Z) _(R,1R), F_(Z) _(R,2L), F_(Z R,2R) in the aboveequations must be limited to values >= 0, since negative forces are notpossible as a result of the elevation of the wheel.

In order to calculate the lateral tire force distribution and themaximum horizontal tire forces, the concept of the effective wheel loadsis introduced. Maximum horizontal tire forces do not increase linearlywith the wheel load, but rather show a slightly degressive behavior.This means that two uniformly loaded wheels together apply a highermaximum horizontal tire force than two non-uniformly loaded wheels, evenif they have the same total vertical forces. There are differentapproaches to model this behavior. Here, an approach is used accordingto R. Orend, Modelling and Control of a vehicle with single-wheelchassis actuators, IFAC Proceedings Volume 38, Issue 1, Pages 79-84,2005, that introduces the effective wheel loads F_(ZT),_(eff,i) asfollows:

$\begin{matrix}{\text{F}_{\text{ZT,eff,i}} = \text{F}_{\text{ZT,i}}\left( {1 + \text{k}_{\text{FZ}}\frac{\text{F}_{\text{ZT,N,i}} - \text{F}_{\text{ZT,l}}}{\text{F}_{\text{ZT,N,l}}}} \right),\quad\text{i} \in \left\{ \text{1L, 1R, 2L, 2R} \right\},} & \text{­­­(67)}\end{matrix}$

wherein 0 ≤ k_(FZ) ≤ 1 refers to an empirical wheel load degressionfactor and F_(ZT,N,i) refers to the nominal wheel load.

If only longitudinal or only lateral tire forces are assumed, therespective peak forces can be modelled as follows:

$\begin{matrix}{\text{F}_{\text{XT,max,i}} = \mspace{6mu}\mu_{\max} \cdot \mspace{6mu}\text{F}_{\text{ZT,eff,i}},} & \text{­­­(68)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YT,max,i}} = \mspace{6mu}\mu_{\max} \cdot \mu_{\text{ql}} \cdot \text{F}_{\text{ZT,eff,i}},\quad\text{i} \in \left\{ \text{1L, 1R, 2L, 2R} \right\},} & \text{­­­(69)}\end{matrix}$

wherein µ_(max) is the potential friction value and µ_(ql) refers to theratio of the lateral traction to the longitudinal traction, whichanisotropically models tire force characteristics.

In order to calculate the horizontal tire forces in the X_(R) direction,the definitions

$\begin{matrix}{\text{F}_{\text{X R,r+f}}: = \text{F}_{\text{X R,r}} + \text{F}_{\text{X R,f}}} & \text{­­­(70)}\end{matrix}$

and (27) are used, in order to form the following:

$\begin{matrix}{\text{F}_{\text{X R,r+f}}: = \text{F}_{\text{X R,r}} + \text{F}_{\text{X R,f}} = - \text{F}_{\text{X R, cg}\text{.}}} & \text{­­­(71)}\end{matrix}$

If the driving and braking torque distributions γ_(d) and γ_(b) areintroduced, which are collectively referred to as y_(d|b), then thefront axle and rear axle forces in the X_(R) direction are given asfollows:

$\begin{matrix}{\text{F}_{\text{XR,f}} = \text{γ}_{\text{d}{|\text{b})}} \cdot \text{F}_{\text{XR,r+f}} = - \text{γ}_{\text{d}{|\text{b})}} \cdot \text{F}_{\text{XR,cg}},} & \text{­­­(72)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{XR,r}} = \left( {1 - \text{γ}_{\text{d}{|\text{b})}}} \right) \cdot \text{F}_{\text{XR,r+f}} = - \left( {1 - \text{γ}_{\text{d}{|\text{b})}}} \right) \cdot \text{F}_{\text{XR,cg}}.} & \text{­­­(73)}\end{matrix}$

Assuming the same left and right driving and braking forces, asmentioned above, this leads to:

$\begin{matrix}{\text{F}_{\text{XR,1L}} = \text{F}_{\text{XR,1R}} = - \frac{\text{γ}_{\text{d}{|\text{b})}}}{2} \cdot \text{F}_{\text{XR,cg}},} & \text{­­­(74)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{XR,2L}} = \text{F}_{\text{XR,2R}} = - \frac{\left( {1 - \text{γ}_{\text{d}{|\text{b})}}} \right)}{2} \cdot \text{F}_{\text{XR,cg}}.} & \text{­­­(75)}\end{matrix}$

In order to calculate the tire forces in the Y_(R) direction, thefollowing function is used, as described in D. Ammon, Modellbildung undSystementwicklung in der Fahrzeugdynamik [Modelling and systemdevelopment in vehicle dynamics], Teubner, Stuttgart, DE, 1997:

$\begin{matrix}\begin{matrix}{\text{F}_{\text{YT,i}} = \frac{\text{α}_{\text{i}}}{\sqrt{\text{α}_{\text{i}}^{2} + \text{s}_{\text{i}}^{2}}}\underset{= \text{μ}_{\max} \cdot \text{μ}_{\text{ql}}}{\underset{︸}{\text{μ}_{\text{Y,max}}}} \cdot \text{F}_{\text{ZT,eff,i}} \cdot \underset{\text{normalized shape function with value sin}{\lbrack{0,1}\rbrack}}{\underset{︸}{\text{g}\left( \frac{\sqrt{\text{α}_{\text{i}}^{2} + \text{s}_{\text{i}}^{2}}}{\text{α}_{\max,\text{N}} \cdot \text{μ}_{\text{γ,}\text{max}}} \right)}},} \\{i\mspace{6mu} \in \left\{ {1L,1R,2L,2R} \right\}.}\end{matrix} & \text{­­­(76)}\end{matrix}$

α_(i) refers to the side-slip angle and s_(i) to the longitudinal slipas defined in D. Ammon, Modellbildung und Systementwicklung in derFahrzeugdynamik, Teubner, Stuttgart, DE, 1997. The normalized shapefunction is saturated for the larger side-slip angle at 1, which is theonly interesting area for further consideration. α_(max,N) is a scalingfactor without further meaning in the present study.

Assuming lateral tire forces F_(Y T,i) with

i  ∈ {1L, 1R, 2L, 2R}

near the limits and therefore in the saturation range of F_(Y) _(T,I),the following assumptions can be made:

$\begin{matrix}{\text{α}_{1\text{R}} \approx \text{α}_{1\text{L}},\mspace{6mu}\mspace{6mu}\text{α}_{2\text{R}} \approx \text{α}_{2\text{L}},\mspace{6mu}\mspace{6mu}\text{s}_{1\text{R}} \approx \text{s}_{1\text{L}},\mspace{6mu}\mspace{6mu}\text{s}_{2\text{R}} \approx \text{s}_{2\text{L}},} & \text{­­­(77)}\end{matrix}$

Using these assumptions and (76), the ratio of left and right tireforces per axle can be estimated as follows:

$\begin{matrix}{\frac{\text{F}_{\text{YR,1L}}}{\text{F}_{\text{YR,1R}}} \approx \frac{\text{F}_{\text{YT,1L}}}{\text{F}_{\text{YT,1R}}} \approx \frac{\text{F}_{\text{ZT,eff,1L}}}{\text{F}_{\text{ZT,eff,1R}}},} & \text{­­­(78)}\end{matrix}$

$\begin{matrix}{\frac{\text{F}_{\text{YR,2L}}}{\text{F}_{\text{YR,2R}}} \approx \frac{\text{F}_{\text{YT,2L}}}{\text{F}_{\text{YT,2R}}} \approx \frac{\text{F}_{\text{ZT,eff,2L}}}{\text{F}_{\text{ZT,eff,2R}}}.} & \text{­­­(79)}\end{matrix}$

This means that the lateral tire force is distributed based on theeffective wheel load ratio.

If tire and axle forces are put in relation in the Y_(R) direction, thenthis results in:

$\begin{matrix}{\text{F}_{\text{Y R,f}} = \text{F}_{\text{Y R, 1L}} + \text{F}_{\text{Y R,1R}},} & \text{­­­(80)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{Y R,f}} = \text{F}_{\text{Y R, 2L}} + \text{F}_{\text{Y R,2R}}.} & \text{­­­(81)}\end{matrix}$

By using (78) and (79), this can be transposed as follows:

$\begin{matrix}{\text{F}_{\text{YR,f}} = \text{F}_{\text{YR,1R}}\left( {\frac{\text{F}_{2\text{T,eff,1L}}}{\text{F}_{2\text{T,eff,1R}}} + 1} \right),} & \text{­­­(82)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,f}} = \text{F}_{\text{YR,1L}}\left( {1 + \frac{\text{F}_{\text{ZT,eff,1R}}}{\text{F}_{\text{ZT,eff,1L}}}} \right),} & \text{­­­(83)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,r}} = \text{F}_{\text{YR,2R}}\left( {\frac{\text{F}_{\text{ZT,eff,2L}}}{\text{F}_{\text{ZT,eff,2R}}} + 1} \right),} & \text{­­­(84)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,r}} = \text{F}_{\text{YR,2L}}\left( {1 + \frac{\text{F}_{\text{ZT,eff,2R}}}{\text{F}_{\text{ZT,eff,2L}}}} \right).} & \text{­­­(85)}\end{matrix}$

Solving according to the tire forces results in:

$\begin{matrix}{\text{F}_{\text{YR,1R}} = \frac{\text{F}_{\text{ZT,eff,1R}}}{\text{F}_{\text{ZT,eff,1L}} + \text{F}_{\text{ZT,eff,1R}}}\text{F}_{\text{YR,f}},} & \text{­­­(86)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,1L}} = \frac{\text{F}_{\text{ZT,eff,1L}}}{\text{F}_{\text{ZT,eff,1L}} + \text{F}_{\text{ZT,eff,1R}}}\text{F}_{\text{YR,f}},} & \text{­­­(87)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,2R}} = \frac{\text{F}_{\text{ZT,eff,2R}}}{\text{F}_{\text{ZT,eff,2L}} + \text{F}_{\text{ZT,eff,2R}}}\text{F}_{\text{YR,r}},} & \text{­­­(88)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,2L}} = \frac{\text{F}_{\text{ZT,eff,2L}}}{\text{F}_{\text{ZT,eff,2L}} + \text{F}_{\text{ZT,eff,2R}}}\text{F}_{\text{YR,r}}.} & \text{­­­(89)}\end{matrix}$

Introducing (25) and (26) results in the tire forces in the Y_(R)direction:

$\begin{matrix}{\text{F}_{\text{YR,1R}} = \frac{\text{F}_{\text{ZT,eff,1R}}}{\text{F}_{\text{ZT,eff,1L}} + \text{F}_{\text{ZT,eff,1R}}}\left( {- \underset{= \frac{1_{\text{h}}}{1}}{\underset{︸}{\left( {1 - \frac{1_{\text{f}}}{1}} \right)}}\text{F}_{\text{YR,cg}} - \frac{1}{1}\text{M}_{\text{ZR,cg}}} \right),} & \text{­­­(90)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,1L}} = \frac{\text{F}_{\text{ZT,eff,1L}}}{\text{F}_{\text{ZT,eff,1L}} + \text{F}_{\text{ZT,eff,1R}}}\left( {- \underset{= \frac{1_{\text{h}}}{1}}{\underset{︸}{\left( {1 - \frac{1_{\text{f}}}{1}} \right)}}\text{F}_{\text{YR,cg}} - \frac{1}{1}\text{M}_{\text{ZR,cg}}} \right),} & \text{­­­(91)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,2R}} = \frac{\text{F}_{\text{ZT,eff,2R}}}{\text{F}_{\text{ZT,eff,2L}} + \text{F}_{\text{ZT,eff,2R}}}\left( {- \frac{1_{\text{f}}}{1}\text{F}_{\text{YR,cg}} + \frac{1}{1}\text{M}_{\text{ZR,cg}}} \right),} & \text{­­­(92)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,2L}} = \frac{\text{F}_{\text{ZT,eff,2L}}}{\text{F}_{\text{ZT,eff,2L}} + \text{F}_{\text{ZT,eff,2R}}}\left( {- \frac{1_{\text{f}}}{1}\text{F}_{\text{YR,cg}} + \frac{1}{1}\text{M}_{\text{ZR,cg}}} \right),} & \text{­­­(93)}\end{matrix}$

Finally, the tire forces in the tire directions X_(T), Y_(T) and Z_(T)are given by:

$\begin{matrix}{\left\lbrack \begin{array}{l}F_{XT,i} \\F_{YT,i} \\F_{ZT,i}\end{array} \right\rbrack = \underset{\text{T}_{RT,i}}{\underset{︸}{\left\lbrack \begin{array}{lll}{\cos\left( \delta_{i} \right)} & {\sin\left( \delta_{i} \right)} & 0 \\{- \sin\left( \delta_{i} \right)} & {\cos\left( \delta_{i} \right)} & 0 \\0 & 0 & 1\end{array} \right\rbrack}}\left\lbrack \begin{array}{l}F_{XR,i} \\F_{YR,i} \\F_{ZR,i}\end{array} \right\rbrack,\mspace{6mu}\mspace{6mu} i\mspace{6mu} \in \left\{ {1L,\mspace{6mu} 1R,\mspace{6mu} 2L,\mspace{6mu} 2R} \right\}.} & \text{­­­(94)}\end{matrix}$

It is hereby possible to check whether the tire forces are within thefriction limit:

$\begin{matrix}{\text{F}_{\text{XT,i}}^{2} + \frac{\text{F}_{\text{YT,i}}^{2}}{\text{μ}_{\text{ql}}^{2}} \leq \mspace{6mu}\text{μ}_{\max}^{2}\text{F}_{\text{ZT,eff,i}}^{2},\mspace{6mu}\mspace{6mu}\text{i} \in \left\{ {1\text{L, 1R, 2L, 2R}} \right\}.} & \text{­­­(95)}\end{matrix}$

FIG. 10 shows a schematic view of a so-called tire force ellipse, inwhich the grip limit for combined forces F_(XT) and F_(YT) isillustrated.

The horizontal target tire forces should be within the friction valuelimits (maximum available potential traction, traction ellipse, see alsoDE 100 50 421 A1) determined by the friction value µ_(max) on everywheel.

If the inequation (95) is fulfilled for every wheel, then one of thethree necessary conditions for drivability is fulfilled, otherwise thetrajectory is not drivable.

Until now, external forces, gravity and inertia have not beenconsidered. Instead, arbitrary force vector components F_(X) _(R,cg),F_(Y) _(R,cg), and F_(Z) _(R,cg), that operate at the center of gravityof the vehicle 1, and arbitrary torque vector components M_(X) _(R,cg),M_(Y) _(R,cg), and M_(Z) _(R,cg) were used as place-holders. In thefollowing, external forces, gravity and inertia are introducedconcerning the stationary influence of transverse incline and gradientas terms, which contribute to F_(X) _(R,cg), F_(Y) _(R,cg), F_(Z R,cg)and M_(X) _(R,cg), M_(Y) _(R,cg), M_(Z) _(R,cg).

Concerning gravity, it follows from FIGS. 1 and 4 that F_(Z,cg,gravity)= -m · g. Therefore:

$\begin{matrix}{\text{F}_{\text{XR,cg,gravity}} = \sin\left( \text{λ} \right) \cdot \text{m} \cdot \text{g,}} & \text{­­­(96)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,cg,gravity}} = - \sin\left( \text{η}_{\text{XR}} \right)\cos\left( \text{λ} \right) \cdot \text{m} \cdot \text{g}\text{=−}\text{sin}\left( \text{η} \right) \cdot \text{m} \cdot \text{g,}} & \text{­­­(97)}\end{matrix}$

$\begin{matrix}\begin{array}{l}{F_{ZR,cg,gravity} = - \cos\left( \eta_{XR} \right)\cos(\lambda) \cdot m \cdot g = - \cos\left( {\arcsin\left( \frac{\sin(\eta)}{\cos(\lambda)} \right)} \right)} \\{\cos(\lambda) \cdot m \cdot g.}\end{array} & \text{­­­(98)}\end{matrix}$

λ is positive when driving downhill.

Concerning translational inertia, it applies that:

In order to express the inertia forces, the accelerations a_(X) _(R,cg),a_(Y) _(R,cg), and a_(Z) _(R,cg) given in road surface plane coordinatesmust be calculated from the desired center of gravity trajectoryaccelerations a_(X,cg) and a_(Y,cg), which are projections in the X-Yplane of the intermediate axle system. Using the assumption ofstationary transverse incline and gradient, the following applies:

$\begin{matrix}{\text{a}_{\text{ZR,cg}} \equiv 0.} & \text{­­­(99)}\end{matrix}$

Therefore, the remaining unknowns are a_(x) _(R,cg) and a_(Y) _(R,cg)and also the movement a_(z,cg) of the vehicle 1, which is not given aspart of the trajectory projections. Using (14), (15) and (99), thefollowing can be determined:

$\begin{matrix}{\text{a}_{\text{XR,cg}} = \cos\left( \text{λ} \right)\text{a}_{\text{X,cg}} - \sin\left( \text{λ} \right)\text{a}_{\text{Z,cg}},} & \text{­­­(100)}\end{matrix}$

$\begin{matrix}\begin{array}{l}{\text{a}_{\text{YR,cg}} = \sin\left( \text{η}_{\text{XR}} \right)\sin\left( \text{λ} \right)\text{a}_{\text{X,cg}} + \cos\left( \text{η}_{\text{XR}} \right)\text{a}_{\text{Y,cg}} +} \\{\sin\left( \text{η}_{\text{XR}} \right)\cos\left( \text{λ} \right)\text{a}_{\text{Z,cg}}.}\end{array} & \text{­­­(101)}\end{matrix}$

$\begin{matrix}\begin{array}{l}{0 = \cos\left( \eta_{XR} \right)\sin(\lambda)a_{X,cg} - \sin\left( \eta_{XR} \right)a_{Y,cg} +} \\{\cos\left( \eta_{XR} \right)\cos(\lambda)a_{Z,cg}.}\end{array} & \text{­­­(102)}\end{matrix}$

$\begin{matrix}\begin{array}{l}{\text{Solving}(102)\text{according to a}_{\text{Z,cg}}\text{results in:}} \\{\text{a}_{\text{Z,cg}} = - \frac{\sin\left( \text{λ} \right)}{\cos\left( \text{λ} \right)}\text{a}_{\text{X,cg}} + \frac{\sin\left( \text{η}_{\text{XR}} \right)}{\cos\left( \text{η}_{\text{XR}} \right)\cos\left( \text{λ} \right)}\text{a}_{\text{Y,cg}}.}\end{array} & \text{­­­(103)}\end{matrix}$

Applying (103) in (100) results in:

$\begin{matrix}\begin{array}{l}{\text{a}_{\text{XR,cg}} =} \\{\cos\left( \text{λ} \right)\text{a}_{\text{X,cg}} - \sin\left( \text{λ} \right)\left( {- \frac{\sin\left( \text{λ} \right)}{\cos\left( \text{λ} \right)}\text{a}_{\text{X,cg}} + \frac{\sin\left( \text{η}_{\text{XR}} \right)}{\cos\left( \text{η}_{\text{XR}} \right)\cos\left( \text{λ} \right)}\text{a}_{\text{Y,cg}}} \right),}\end{array} & \text{­­­(104)}\end{matrix}$

$\begin{matrix}{\text{a}_{\text{XR,cg}} = \frac{1}{\cos\left( \text{λ} \right)}\underset{= 1}{\underset{︸}{\left( {\cos^{2}\left( \text{λ} \right) + \sin^{2}\left( \text{λ} \right)} \right)}}\text{a}_{\text{X,cg}} - \frac{\sin\left( \text{η}_{\text{XR}} \right)\sin\left( \text{λ} \right)}{\cos\left( \text{η}_{\text{XR}} \right)\cos\left( \text{λ} \right)}\text{a}_{\text{Y,cg}},} & \text{­­­(105)}\end{matrix}$

$\begin{matrix}{\text{a}_{\text{XR,cg}} = \frac{1}{\cos\left( \text{λ} \right)}\text{a}_{\text{X,cg}} - \frac{\sin\left( \text{η}_{\text{XR}} \right)\sin\left( \text{λ} \right)}{\cos\left( \text{η}_{\text{XR}} \right)\cos\left( \text{λ} \right)}\text{a}_{\text{Y,cg}}.} & \text{­­­(106)}\end{matrix}$

Applying (103) in (101) results in:

$\begin{matrix}{\text{a}_{\text{YR,cg}} = \sin\left( \text{η}_{\text{XR}} \right)\sin\left( \text{λ} \right)\text{a}_{\text{X,cg}} + \cos\left( \text{η}_{\text{XR}} \right)\text{a}_{\text{Y,cg}}} & \text{­­­(107)}\end{matrix}$

$\begin{matrix}{+ \sin\left( \text{η}_{\text{XR}} \right)\cos\left( \text{λ} \right)\mspace{6mu}\left( {- \frac{\sin\left( \text{λ} \right)}{\cos\left( \text{λ} \right)}\text{a}_{\text{X,cg}} + \frac{\sin\left( \text{η}_{\text{XR}} \right)}{\cos\left( \text{η}_{\text{XR}} \right)\cos\left( \text{λ} \right)}\text{a}_{\text{Y,cg}}} \right),} & \text{­­­(108)}\end{matrix}$

$\begin{matrix}{\text{a}_{\text{YR,cg}} = \left( {\cos\left( \text{η}_{\text{XR}} \right) + \frac{\sin^{2}\left( \text{η}_{\text{XR}} \right)}{\cos\left( \text{η}_{\text{XR}} \right)}} \right)\text{a}_{\text{Y,cg}},} & \text{­­­(109)}\end{matrix}$

$\begin{matrix}{a_{YR,cg} = \frac{1}{cos\left( \eta_{XR} \right)}a_{y,cg}.} & \text{­­­(110)}\end{matrix}$

Finally, the translational inertia force vector components can bedetected with (99), (106) and (110) as follows:

$\begin{matrix}\begin{array}{l}{\text{F}_{\text{XR,cg,inert,trans}} = - \text{m} \cdot \text{a}_{\text{XR,cg}} = - \frac{\text{m}}{\cos\left( \text{λ} \right)}\text{a}_{\text{X,cg}} +} \\{\text{m}\frac{\sin\left( \text{η}_{\text{XR}} \right)\sin\left( \text{λ} \right)}{\cos\left( \text{η}_{\text{XR}} \right)\cos\left( \text{λ} \right)}\text{a}_{\text{Y,cg}},}\end{array} & \text{­­­(111)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{YR,cg,inert,trans}} = - \text{m} \cdot \text{a}_{\text{YR,cg}} = - \frac{\text{m}}{\cos\left( \text{η}_{\text{XR}} \right)}\text{a}_{\text{Y,cg}},} & \text{­­­(112)}\end{matrix}$

$\begin{matrix}{\text{F}_{\text{ZR,cg,inert,trans}} = - \text{m} \cdot \text{a}_{\text{ZR,cg}} = 0.} & \text{­­­(113)}\end{matrix}$

Concerning the rotational inertia, it applies that: In order tocalculate the rotational torque M_(X R,cg,inert,rot),M_(Y R,cg,inert,rot), and M_(Z R,cg,inert,rot) that is related to therotational inertia, the rotational accelerations ω_(ẋR), ω_(ẏR), ω_(żR),and of the vehicle 1, which are given in road surface plane coordinates,must be expressed in terms of the desired trajectory of the center ofgravity.

Using D.T. Greenwood, Principles of Dynamics, Prentice Hall, UpperSaddle River, US-NJ, 2nd edition, 1988, p. 406, the relationship betweenEuler angles and angular speeds is notated as in D.T. Greenwood,Principles of Dynamics, Prentice Hall, Upper Saddle River, US-NJ, 2^(nd)edition, 1988:

$\begin{matrix}{\text{ω}_{\text{x}} = \text{ϕ−ψ}\text{sin}\left( \text{θ} \right)} & \text{­­­(114)}\end{matrix}$

$\begin{matrix}{\text{ω}_{\text{y}} = \text{θ}\text{cos}\left( \text{ϕ} \right) + \text{ψ}\text{cos}\left( \text{θ} \right)\sin\left( \text{ϕ} \right)} & \text{­­­(115)}\end{matrix}$

$\begin{matrix}{\text{ω}_{\text{z}} = \text{ψ}\text{cos}\left( \text{θ} \right)\cos\left( \text{ϕ} \right) - \text{θ}\text{sin}\left( \text{ϕ} \right)} & \text{­­­(116)}\end{matrix}$

Using the equivalents

$\begin{matrix}{\text{ω}_{\text{x}} ≙ \text{ω}_{\text{XR}},\text{ω}_{y} ≙ \text{ω}_{\text{YR}},\text{ω}_{\text{z}} ≙ \text{ω}_{\text{ZR}},} & \text{­­­(117)}\end{matrix}$

$\begin{matrix}{\text{ϕ} ≙ \text{η}_{\text{XR}},\text{θ} ≙ \text{λ,ψ} ≙ \text{ψ,}} & \text{­­­(118)}\end{matrix}$

this translates into the notation of the present document:

$\begin{matrix}{\text{ω}_{\text{XR}} = {\overset{˙}{\text{η}}}_{\text{XR}} - \text{ψ}\text{sin}\left( \text{λ} \right)} & \text{­­­(119)}\end{matrix}$

$\begin{matrix}{\text{ω}_{\text{YR}} = \overset{˙}{\text{λ}}\text{cos}\left( \text{η}_{\text{XR}} \right)\mspace{6mu} + \mspace{6mu}\overset{˙}{\text{ψ}}\text{cos}\left( \text{λ} \right)\sin\left( \text{η}_{\text{XR}} \right)} & \text{­­­(120)}\end{matrix}$

$\begin{matrix}{\omega_{ZR} = \overset{˙}{\psi}\text{cos}(\lambda)\cos\left( \eta_{XR} \right) - \overset{˙}{\lambda}\text{sin}\left( \eta_{XR} \right)} & \text{­­­(121)}\end{matrix}$

Assuming steady conditions of slowly changing transverse incline andgradients, i.e.

${\overset{˙}{\eta}}_{XR} \approx 0\prime$

and

$\overset{˙}{\lambda} \approx 0$

0, this leads to:

$\begin{matrix}{\text{ω}_{\text{XR}} = - \text{ψ}\text{sin}\left( \text{λ} \right)} & \text{­­­(122)}\end{matrix}$

$\begin{matrix}{\text{ω}_{\text{YR}} = \overset{˙}{\text{ψ}}\text{cos}\left( \text{λ} \right)\sin\left( \text{η}_{\text{XR}} \right)} & \text{­­­(123)}\end{matrix}$

$\begin{matrix}{\text{ω}_{\text{ZR}} = \overset{˙}{\text{ψ}}\text{cos}\left( \text{λ} \right)\cos\left( \text{η}_{\text{XR}} \right)} & \text{­­­(124)}\end{matrix}$

Taking into consideration the following relationship between the angularspeed derivations

and

if they are considered in an inert context and a moving context (seeD.T. Greenwood, Principles of Dynamics, Prentice Hall, Upper SaddleRiver, US-NJ, 2^(nd) edition, 1988, p. 392):

$\begin{matrix}\begin{array}{l}{\overset{˙}{\text{ω}}\text{=}\underset{\text{as seen from moving frame}}{\underset{︸}{\left( \overset{˙}{\text{ω}} \right)_{\text{r}}}} + \underset{= 0}{\underset{︸}{\text{ω} \times \text{ω,}}}} \\{= \left( \overset{˙}{\text{ω}} \right)_{r},}\end{array} & \text{­­­(125)}\end{matrix}$

the angular accelerations can be found by calculating the derivations of(122) to (124) using

${\overset{˙}{\eta}}_{XR} \approx 0\prime$

and

$\overset{˙}{\lambda} \approx 0:$

$\begin{matrix}{{\overset{˙}{\text{ω}}}_{\text{XR}} = - \overset{¨}{\text{ψ}}\text{sin}\left( \text{λ} \right)} & \text{­­­(126)}\end{matrix}$

$\begin{matrix}{{\overset{˙}{\text{ω}}}_{\text{YR}} = \overset{¨}{\text{ψ}}\text{cos}\left( \text{λ} \right)\sin\left( \text{η}_{\text{XR}} \right)} & \text{­­­(127)}\end{matrix}$

$\begin{matrix}{{\overset{˙}{\text{ω}}}_{\text{ZR}} = \overset{¨}{\text{ψ}}\text{cos}\left( \text{λ} \right)\cos\left( \text{η}_{\text{XR}} \right)} & \text{­­­(128)}\end{matrix}$

With Euler’s equation of movement (D.T. Greenwood, Principles ofDynamics, Prentice Hall, Upper Saddle River, US-NJ, 2^(nd) edition,1988, p. 392) and also (122) to (124) and (126) to (128), it is possibleto calculate the rotation torque resulting from the inertia:

$\begin{matrix}{\text{M}_{\text{XR,cg,inert,rot}} = - \text{J}_{\text{xx}}{\overset{˙}{\text{ω}}}_{\text{XR}} - \left( {\text{J}_{\text{zz}} - \text{J}_{\text{yy}}} \right)\text{ω}_{\text{YR}}\text{ω}_{\text{ZR}}} & \text{­­­(129)}\end{matrix}$

$\begin{matrix}{= \mspace{6mu} + \text{J}_{\text{xx}}\overset{¨}{\text{ψ}}\text{sin}\left( \text{λ} \right) - \left( {\text{J}_{\text{zz}} - \text{J}_{\text{yy}}} \right){\overset{˙}{\text{ψ}}}^{2}\cos^{2}\left( \text{λ} \right)\sin\left( \text{η}_{\text{XR}} \right)\cos\left( \text{η}_{\text{XR}} \right)} & \text{­­­(130)}\end{matrix}$

$\begin{matrix}{\text{M}_{\text{YR,cg,inert,rot}} = - \text{J}_{\text{yy}}{\overset{˙}{\text{ω}}}_{\text{YR}} - \left( {\text{J}_{\text{xx}} - \text{J}_{\text{zz}}} \right)\text{ω}_{\text{ZR}}\text{ω}_{\text{XR}}} & \text{­­­(131)}\end{matrix}$

$\begin{matrix}{= \mspace{6mu} - \text{J}_{\text{yy}}\overset{¨}{\text{ψ}}\text{cos}\left( \text{λ} \right)\sin\left( \text{η}_{\text{XR}} \right) + \left( {\text{J}_{\text{xx}} - \text{J}_{\text{zz}}} \right){\overset{˙}{\text{ψ}}}^{2}\sin\left( \text{λ} \right)\cos\left( \text{λ} \right)\cos\left( \text{η}_{\text{XR}} \right)} & \text{­­­(132)}\end{matrix}$

$\begin{matrix}{\text{M}_{\text{ZR,cg,inert,rot}} = - \text{J}_{\text{zz}}{\overset{˙}{\text{ω}}}_{\text{ZR}} - \left( {\text{J}_{\text{yy}} - \text{J}_{\text{xx}}} \right)\text{ω}_{\text{XR}}\text{ω}_{\text{YR}}} & \text{­­­(133)}\end{matrix}$

$\begin{matrix}{= \mspace{6mu} - \text{J}_{\text{zz}}\overset{¨}{\text{ψ}}\text{cos}\left( \text{λ} \right)\cos\left( \text{η}_{\text{XR}} \right) + \left( {\text{J}_{\text{yy}} - \text{J}_{\text{xx}}} \right){\overset{˙}{\text{ψ}}}^{2}\sin\left( \text{λ} \right)\cos\left( \text{λ} \right)\sin\left( \text{η}_{\text{XR}} \right)} & \text{­­­(134)}\end{matrix}$

Regarding the aerodynamic drag and lift, it is possible to express thecontributions of aerodynamic drag (M. Mitschke, Dynamik derKraftfahrzeuge [Motor vehicle dynamics], Springer, Berlin, DE, 5^(th)edition, 2014.) to the arbitrary forces and torques operating at thecenter of gravity of the vehicle 1 by introducing the height of thetraction lever arm h_(d), which describes the resulting height of thedrag forces above the ground:

$\begin{matrix}{\text{F}_{\text{XR,cg,drag}} = - \text{sgn}\left( \text{V}_{\text{XR}} \right)\text{C}_{\text{d}}\frac{\text{ρ}_{\text{a}}}{2}\text{A}_{\text{a}}\text{V}_{\text{XR}}^{2},} & \text{­­­(135)}\end{matrix}$

$\begin{matrix}{\text{M}_{\text{YR,cg,drag}} = \left( {\text{h}_{\text{d}} - \text{h}_{\text{cg}}} \right)\text{F}_{\text{XR,cg,drag}}} & \text{­­­(136)}\end{matrix}$

$\begin{matrix}{= - \left( {h_{d} - h_{cg}} \right)\text{sgn}\left( v_{XR} \right)C_{d}\frac{\rho_{a}}{2}A_{a}v_{XR}^{2}.} & \text{­­­(137)}\end{matrix}$

C_(d) hereby refers to the aerodynamic drag coefficients, A_(a) to theaerodynamic surfaces and p_(a) to the air density.

In a similar way, the aerodynamic lift (M. Mitschke, Dynamik derKraftfahrzeuge, Springer, Berlin, DE, 5th edition, 2014.) is describedby:

$\begin{matrix}{\text{F}_{\text{ZR,cg,lift}} = \left( {\text{C}_{1,\text{f}} + \text{C}_{1,\text{r}}} \right)\frac{\text{ρ}_{\text{a}}}{2}\text{A}_{\text{a}}\text{V}_{\text{XR}}^{2},} & \text{­­­(138)}\end{matrix}$

$\begin{matrix}{\text{M}_{\text{YR,cg,lift}} = \left( {- 1_{\text{l}}\text{C}_{1,\text{f}} + \left( {1 - 1_{\text{f}}} \right)\text{C}_{1,\text{r}}} \right)\frac{\text{ρ}_{\text{a}}}{2}\text{A}_{\text{a}}\text{v}_{\text{XR}}^{2},} & \text{­­­(139)}\end{matrix}$

C_(l,f) and C_(l,r) hereby refer to front and rear aerodynamic liftcoefficients.

Regarding the quasi-steady-state steering system model, it applies that:In order to assess the physical feasibility (=drivability) of atrajectory, steering force limitations are a further important aspect.Considering the tire forces, which are discussed above in the context ofthe equations (70) to (95), limitations of the steering system and ofits power are discussed in the following. The assumptions of the modelare met as discussed above.

The assessment of the drivability of trajectories is especially relevantat relatively high speeds. In this case, limitations of the steeringforce and steering rate must be considered in order to ensure theselection of drivable trajectories. Slow cornering and parking maneuverscan easily be handled by rescheduling, in the case that variations occurbetween desired and actual steering actuator movements.

For higher speeds and smaller steering angles, W. Matschinsky,Radführungen der Straβenfahrzeuge [Road vehicle wheel control],Springer, Berlin, DE, 3rd edition, 2007 states that the steering torqueis dominated by lateral tire forces F_(Y) _(T,1L), F_(Y T,1R), kinematiclag n_(k) and tire lag n_(p). The kinematic lag defined by steeringkinematics typically changes with the vertical wheel hub. The tire lagchanges with the longitudinal tire slip and with the side-slip angle ofthe tire. The nearer the tire is to the slip conditions, the smaller thetire lag becomes.

When there are equal left and right driving/ braking forces, asconsidered above, the influences of F_(XT,1L) and F_(XT,1R) on thesteering force cancel each other out. Therefore, the necessary steeringforce is dominated by the contribution of the lateral forces F_(Y T,1L)and F_(Y T,1R) to the return torque. This leads to:

$\begin{matrix}{P_{s} = \underset{F_{YT,1L}\text{part of self aligning torque}}{\underset{︸}{\left( {n_{k,1L} + n_{p,1L}} \right)F_{YT,1L}}}{\overset{˙}{\delta}}_{1L} + \underset{F_{YT,1R}\text{part of self aligning torque}}{\underset{︸}{\left( {n_{k,1R} + n_{p,1R}} \right)F_{YT,1R}}}{\overset{˙}{\delta}}_{1R}.} & \text{­­­(140)}\end{matrix}$

Since kinematic lag n_(k) and tire lag n_(p) depend on the verticalwheel hub and slipping, a quasi-steady-state model, that does not takeinto consideration the chassis movement, stimuli from the road surfaceand slipping, is not capable of providing exact values for n_(k) andn_(p). It is, however, possible to calculate an upper limit P_(s),_(max)≥P_(s). With the definitions of the maximum kinematic lag n_(k)and tire lag n_(p) and the average steering rate

_(1,m) as follows:

$\begin{matrix}{\text{n}_{\text{k,max}}:\mspace{6mu} = \max\left( {\text{n}_{\text{k,1L}}(\ldots),\text{n}_{\text{k,1R}}(\ldots)} \right),} & \text{­­­(141)}\end{matrix}$

$\begin{matrix}{\text{n}_{\text{p,max}}:\mspace{6mu} = \max\left( {\text{n}_{\text{p,1L}}(\ldots),\text{n}_{\text{p,1R}}(\ldots)} \right),} & \text{­­­(142)}\end{matrix}$

$\begin{matrix}{{\overset{˙}{\text{δ}}}_{1,\text{m}}:\mspace{6mu} = \mspace{6mu}\frac{\text{δ}_{1\text{L}} + \text{δ}_{1\text{R}}}{2},} & \text{­­­(143)}\end{matrix}$

the upper limit is given by:

$\begin{matrix}{P_{s,max} = \left( {n_{k,max} + n_{p,max}} \right)\left( {F_{YT,1L} + F_{YT,1R}} \right){\overset{˙}{\delta}}_{1,m}} & \text{­­­(144)}\end{matrix}$

A further of the three necessary conditions of the drivability test thusneeds to be tested, whether P_(s,max) is below the presently availablepower of the electric power steering (EPS) P_(EPS). If the inequationP_(s,max) ≤ P_(EPS) is fulfilled, then this necessary condition fordrivability is fulfilled, otherwise the trajectory is not drivable.

Concerning the quasi-steady-state drive train model, it applies that:The core idea in order to assess the drivability limitations in thedrive train is to compare the available traction and wheel power withthe available force and power on the wheels (identifying delivery).Details of the definition of these terms are mentioned in M. Mitschke,Dynamik der Kraftfahrzeuge, Springer, Berlin, DE, 5th edition, 2014.

The traction requirement F_(traction,demand) as defined in (145) is usedin order to differentiate the following cases:

-   1. Driving: F_(traction,demand) ≥ 0-   2. Engine braking: F_(traction,hyd) brake threshold <    F_(traction,demand) < 0-   3. Engine braking and hydraulic braking: F_(traction,demand) ≤    F_(traction,hyd) _(brake) threshold

In order to model the required and available traction and wheel power,the following assumptions of the model are made.

The following aspects are taken into consideration:

-   acceleration resistance of the chassis and vehicle body-   acceleration resistance of the drive train-   gradient resistance-   two wheel drive, four wheel drive, four wheel drive with independent    motors-   power losses due to wheel slippage-   power losses of the drive train-   electric actuator or internal combustion engine-   manual gearbox in steady conditions (power and force envelope curve    for all gears)-   roll resistance (without speed dependence)-   degradation of the electric actuator (starting power, hourly    performance, continuous power)-   aerodynamic drag

The following aspects are disregarded:

-   gear shifting strategy and shifting process. For example, it is    assumed that the gearbox controller chooses an appropriate gear in    order to provide the required wheel power and traction, and that    there is no significant loss of friction during the switching.-   potential friction, i.e., the available traction only models the    engine limits, friction limits are already taken into consideration    with the tire force in the longitudinal direction.-   electrical reinforcement of the internal combustion engine-   dual braking between the motor/electric actuator and hydraulic brake    is provided, however is not yet modelled in detail.

Case 1: Driving

In the following considerations, an individual internal combustionengine or an individual electric actuator is provided. The approach canhowever be easily extended, in that the drive torque is distributedacross several axles by means of the drive torque distribution factor.

According to M. Mitschke, Dynamik der Kraftfahrzeuge, Springer, Berlin,DE, 5th edition, 2014, the total traction requirement is given by:

$\begin{matrix}{\text{F}_{\text{traction,demand}} = \text{f}_{\text{R}} \cdot \text{m} \cdot \text{g}\text{−}\text{F}_{\text{XR,cg}} + \left( {\text{λ}_{\text{m}} - 1} \right) \cdot \text{m} \cdot \text{a}_{\text{XR,cg}},} & \text{­­­(145)}\end{matrix}$

with

-   f_(R) · m · g roll resistance-   F_(X) _(R,cg): total of the gradient resistance (96), chassis and    vehicle body inertia (111) and of the aerodynamic drag (135),-   (λ_(m) - 1) · m · a_(X) _(R,cg): acceleration resistance of the    drive train with the mass factor λ_(m) for rotatable masses.

The wheel power requirement without tire slippage loss is simply givenby:

$\begin{matrix}{\text{P}_{\text{traction}\text{.demand}\text{.}{\text{w}/\text{o slip}}} = \text{F}_{\text{traction}\text{.demand}} \cdot \text{V}_{\text{X R,cg}}} & \text{­­­(146)}\end{matrix}$

The wheel power with tire slip is given by:

$\begin{matrix}{\text{P}_{\text{traction,demand}} = \frac{1}{\text{η}_{\text{tire}}} \cdot \text{F}_{\text{traction,demand}} \cdot \text{V}_{\text{XR,cg}},\text{with}\text{η}_{\text{T}}: = \frac{\text{r}_{\text{d}}\left( {1 - \text{s}} \right)}{\text{r}_{\text{s}}},} & \text{­­­(147)}\end{matrix}$

and the following definitions:

-   η_(T) : describes tire efficiency, the losses as a result of    longitudinal slip,-   s: average slip for the driven axle, assuming e.g. s ≈ s_(2L) ≈    s_(2R)-   r_(d): the dynamic wheel radius, with-   $r_{d} = \frac{u}{2_{\text{N}}}$-   and U as the distance that was covered on the ground by the moving    wheel-   r_(s): the static wheel radius, i.e. the distance from the center of    the wheel to the road surface.

The following differences should be noted:

-   2π * r_(d): distance that was covered on the ground by the slip-free    wheel,-   2π * r_(d)(1 - s): distance that was covered on the ground by the    slipping wheel,-   2π * r_(s): distance that was covered on the ground by a point on    the steady wheel with the radius r_(s).

For the quasi-steady-state vehicle model as described above, thelongitudinal slip is unknown. When anti-slip control is switched on, thelongitudinal slip is, in the worst case scenario, approximately thecritical slip s_(c) (M. Mitschke, Dynamik der Kraftfahrzeuge, Springer,Berlin, DE, 5th edition, 2014), typically around 10%. Therefore, a tireefficiency at the critical slippage can be calculated as follows:

$\begin{matrix}{\eta_{T,c} = \frac{r_{d}\left( {1 - s_{c}} \right)}{r_{N}}} & \text{­­­(148)}\end{matrix}$

This leads to an upper limit approximation for P_(traction,demand). Amore detailed assessment could be achieved by interpolating η_(T)between values of 1 to η_(T), _(c), for example based on the frictionlimit utilization:

$\begin{matrix}{\text{friction limit utilization:}\text{=}\frac{1}{2}\left( {\frac{\text{F}_{\text{XT,2L}}^{2} + \frac{\text{F}_{\text{YT,2L}}^{2}}{\text{μ}_{\text{q1}}^{2}}}{\text{μ}_{\max}^{2}\text{F}_{2\text{T,eff,2L}}^{2}} + \frac{\text{F}_{\text{XT,2R}}^{2} + \frac{\text{F}_{\text{YT,2R}}^{2}}{\text{μ}_{\text{q1}}^{2}}}{\text{μ}_{\max}^{2}\text{F}_{2\text{T,eff,2R}}^{2}}} \right)} & \text{­­­(149)}\end{matrix}$

FIG. 9 shows schematic views of look-up tables LUT (characteristicdelivery fields). F_(traction,supply) and P_(traction,supply) are givenas look-up tables (characteristic delivery field) for an electric enginewith a fixed gear transmission (continuous line) and for an internalcombustion engine with a manual gearbox (dotted line). The supplylook-up tables describe the traction and power that are available on theaxle. This means they already take into consideration the drive trainlosses. In the case of a manual gearbox, the outer envelope curve isused for all gears. The look-up tables can be scaled according to thedegradation DEG.

In order to determine drivability, F_(traction,demand) andP_(traction,demand) are compared to the envelope curves ofF_(traction,supply) and P_(traction,supply), which are given as a(characteristic delivery field), as is illustrated in FIG. 9 . Thelook-up tables already take into consideration the drive train losses,which are assumed to be modelled as constant efficiency factors for eachgear. The degradation can be taken into consideration in that areduction factor is used, which, for example, is based on enginetemperature, power supply, etc.

If the inequations F_(traction,supply) ≥ F_(traction,demand) andP_(traction,supply) ≥ P_(traction,demand) are fulfilled, then the thirdnecessary condition for drivability is fulfilled, otherwise thetrajectory is not drivable.

Case 2: Engine Braking

A small negative traction requirement F_(traction,demand) is realized,exclusively using engine braking. For a two-wheel-drive vehicle 1, thismeans that braking tire forces only occur on the front tires or the reartires. Especially on surfaces with low friction, this can cause lockingtires if no further measures are taken. Therefore, the control for theelectronic stability program is needed in order to prevent lockingtires. In this case, the hydraulic brakes are activated and case 2 skipsover to case 3. This means that there is no restriction of the tractionand traction supply besides the friction limits, wherein the frictionlimits are already taken into consideration in the longitudinal tireforce (see above).

Case 3: Engine Braking and Hydraulic Braking

In this case, there is no restriction of the traction and tractionsupply besides the friction limits, wherein the friction limits arealready taken into consideration in the longitudinal tire force (seeabove).

The following table 1 gives an overview of the vehicle parameters.

Parameter Unit Description m kg overall vehicle mass I m wheelbase I_(f)m distance between the center of gravity and the front axle I_(r) mdistance between the center of gravity and the rear axle (I_(r) = I -I_(f)) h_(cg) m height of the center of gravity h_(rf) m height of thefront roll center h_(rr) m height of the rear roll center h_(rcg) mheight of the roll center on the center of gravity (h_(rcg) = h_(rr) +(h_(rf) - h_(rr))(1 - I_(f)/I)) b_(f) m front track width b_(r) m reartrack width K - roll torque distribution between front and rear axle (1:only front, 0: only rear) Y_(d) - drive torque distribution betweenfront and rear axle (1: only front, 0: only rear) Y_(b) - brake torquedistribution between front and rear axle (1: only front, 0: only rear)k_(FZ) - wheel load digression factor (0 ≤ k_(FZ) << 1) F_(ZT,N,1R,) . .., F_(ZT,N,2L) N nominal wheel load in static load conditions µ_(max) -potential friction µ_(ql) - ratio of the lateral to the longitudinalgrip (Idea: µ² F_(z) ² = F_(x) ² + F_(y) ² /µ² _(ql)I) n_(k,max) mmaximum kinematic lag n_(p,max) m maximum tire lag J_(xx), J_(yy),J_(zz) kg m² inertia torque of the vehicle 1 around the vehicle axles atthe center of gravity C_(d) - aerodynamic drag coefficient Aa m²aerodynamic surface h_(d) m height of the traction lever arm C_(l,f) -aerodynamic lift coefficient, front C_(l,r) - aerodynamic liftcoefficient, rear ρ_(a) kg/m3 air density λ_(m) - mass factor forrotatable masses in the drive train (depending on the gear) f_(R) - rollresistance factor r_(s) m static wheel radius (distance from the centerof the wheel to the road surface) r_(d) m dynamic wheel radiusU=2πr_(d), U: distance that was covered by the moving wheel η_(T,c) -power efficiency of the tire with the critical slip

FIG. 11 demonstrates the functional principle of the drivability testingFP for autonomous vehicles 1.

Input parameters are:

-   Vehicle trajectories to have their drivability examined.-   The slope λ and transverse incline η of the street in front of the    vehicle 1, for example from sensor or map data-   The potential friction value µ_(max) of the street, for example from    sensor or map data-   Information about the degradation of the steering and drive actuator    (available power/nominal power, available torque/ nominal torque)

Output parameters are:

-   The assessment of the drivability (yes/ no)-   Optionally, a quantitative assessment of the exploitation of the    potential friction value

The technical signal processing occurs as follows: In a first step S1, aconversion of the requirements into target trajectory parameters (targetposition, target acceleration, target speed, target camber, targetchange in camber, etc.) as well as a conversion of the gradient andtransverse incline into requirements in vehicle parameters (targettraction, target tractive power, target steering power, target tireforces) is carried out. The conversion thereby occurs by means of aninverse vehicle model, for example the above-describedquasi-steady-state model.

In a second step S2, a comparison of the requirements in vehicleparameters with the limits predetermined by the vehicle 1 and road iscarried out, i.e.,:

-   a drive train limit check DTLC of whether the target traction and    target tractive power are below the limits given by the    characteristic traction curve and characteristic tractive power    curve at the target speed,-   a steering limit check SLC of whether the target steering power is    below the power limit of the steering, and-   a friction value limit check FLC of whether horizontal target tire    forces are within the friction value limits (Kamm circle)

The limits in the second step can thereby be adjusted to the currentavailable power with the assistance of information about degradation.

The result of the drivability test is finally aggregated and isdescribed by a binary statement of “drivable/not drivable” for everytrajectory analyzed. A more detailed evaluation of the drivability isconceivable by means of the calculation of the overshooting of the limitor of the still-available distance from the limit in percent. A possibleexample of feedback could appear as the following: tire force too highon the front left, 120% of the potential friction value used. Thisextension offers the advantage of providing more detailed auxiliaryinformation for planning trajectories.

Although the invention has been illustrated and described in detail byway of preferred embodiments, the invention is not limited by theexamples disclosed, and other variations can be derived from these bythe person skilled in the art without leaving the scope of theinvention. It is therefore clear that there is a plurality of possiblevariations. It is also clear that embodiments stated by way of exampleare only really examples that are not to be seen as limiting the scope,application possibilities or configuration of the invention in any way.In fact, the preceding description and the description of the figuresenable the person skilled in the art to implement the exemplaryembodiments in concrete manner, wherein, with the knowledge of thedisclosed inventive concept, the person skilled in the art is able toundertake various changes, for example, with regard to the functioningor arrangement of individual elements stated in an exemplary embodimentwithout leaving the scope of the invention, which is defined by theclaims and their legal equivalents, such as further explanations in thedescription.

Reference numerals list 1 vehicle 1L left front wheel 1R right frontwheel 2L left rear wheel 2R right rear wheel b_(f) track width of thefront axle DEG degradation DTLC drive train limit check FLC frictionvalue limit check FRC front roll center, roll center of the front axleFP drivability testing F_(traction,supply) available traction F_(X)_(R,cg,) F_(Y) _(R,cg,) F_(Z) _(R,cg) force vector components F_(X R,f,)F_(X R,r,) F_(Z R,r) axial forces F_(X) _(R,R1), F_(Z R,R1), tire forcesF_(X R,L1), F_(Z R,L1), tire forces F_(X) _(R,R2), F_(Z R,R2), tireforces F_(X) _(R,L2), F_(Z R,L2), tire forces F_(Y) _(R,1L), F_(Z R,1L),tire forces F_(Y) _(R,1R), F_(Z R,1R) tire forces F_(X T), F_(Y T)horizontal target tire forces F_(Y R,1), F_(Z R,1) chassis reactionforces F_(z,eff) effective tire load h_(cg) height of the center ofgravity h_(rcg) height of the roll center at the x_(R) position of thecenter of gravity h_(rf) height of the roll center of the front axle Iwheelbase I_(f) space between the center of gravity and the front axleLUT look-up table LTM long-term storage M_(X) _(R,cg,) M_(Y) _(R,cg,)M_(Z R,cg) torque vector components M_(XR,1) chassis reaction torqueP_(traction,supply) available tractive power RA roll axle RC_(cg) rollcenter at the x_(R) position of the center of gravity RRC rear rollcenter S1 first step S2 second step SLC steering limit check V_(XR,)_(cg) longitudinal speed in the center of gravity X, Y, Z intermediateaxle system X_(E), Y_(E), Z_(E) fixed axle system X_(R), Y_(R), Z_(R)road surface plane axle system, road axle system X_(T,1L), Y_(T,1L),Z_(T,1) _(L) tire axle system X_(V), Y_(V),Z_(V) vehicle axle system X′,Y′, Z′ auxiliary axle system δ_(1L), δ₁ _(R) steering angle η roadsurface-level camber angle, transverse incline n_(X) _(R) angle θ angleof incline λ road surface slope angle, road surface-level slope, slopeµ_(max) potential friction value, friction value µql ratio of thelateral grip to the longitudinal grip φ roll angle φV vehicle roll angleΨ yaw angle

1-10. (canceled)
 11. A method for checking the suitability of a targettrajectory for trajectory control of a vehicle, the method comprising:detecting or estimating a slope, a transverse incline, and a frictionvalue of a road surface along the target trajectory using sensors or mapdata, wherein the target trajectory contains route information about thepath of a route to be driven and momentum information about momentumwith which the route should be driven; calculating target valuesrequired for driving the target trajectory of target traction, targettractive power, target steering power, and horizontal target tire forceson individual wheels of the vehicle based on model equations of thevehicle from the route information and momentum information about thetarget trajectory and from the detected slope and transverse incline,wherein the target trajectory is assessed as suitable for the trajectorycontrol if: the target traction and target tractive power are below apredetermined characteristic traction curve or characteristic tractivepower curve, the target steering power is below a predetermined powerlimit of the steering, and the horizontal target tire forces on eachwheel are within the friction value limits determined by the frictionvalue, wherein the target trajectory is otherwise assessed to beunsuitable.
 12. The method of claim 11, wherein the method is performedon every target trajectory from a predetermined plurality of targettrajectories.
 13. The method of claim 11, wherein every targettrajectory assessed as unsuitable is rejected as invalid.
 14. The methodof claim 11, wherein the predetermined characteristic traction curve andthe characteristic tractive power curve are obtained from respectivelook-up tables.
 15. The method of claim 14, wherein the predeterminedcharacteristic traction curve and the characteristic tractive powercurve take into consideration occurrences of degradation of a drivetrain of the vehicle.
 16. The method of claim 11, further comprising:performing a quantitative assessment of exploitation of a potentialfriction value.
 17. The method of claim 11, wherein occurrences ofdegradation of a steering actuator are taken into consideration.
 18. Themethod of claim 11, wherein the model equations of the vehicle rely on aquasi-steady-state modelling approach.
 19. The method of claim 11,wherein in a first step, parameters of the target trajectory, the slope,and the transverse incline are converted into vehicle parameters usingthe model equations.
 20. A device, configured to check the suitabilityof a target trajectory for trajectory control of a vehicle, wherein thedevice is configured to: detect or estimate a slope, a transverseincline, and a friction value of a road surface along the targettrajectory using sensors or map data, wherein the target trajectorycontains route information about the path of a route to be driven andmomentum information about momentum with which the route should bedriven; calculate target values required for driving the targettrajectory of target traction, target tractive power, target steeringpower, and horizontal target tire forces on individual wheels of thevehicle based on model equations of the vehicle from the routeinformation and momentum information about the target trajectory andfrom the detected slope and transverse incline, wherein the targettrajectory is assessed as suitable for the trajectory control if: thetarget traction and target tractive power are below a predeterminedcharacteristic traction curve or characteristic tractive power curve,the target steering power is below a predetermined power limit of thesteering, and the horizontal target tire forces on each wheel are withinthe friction value limits determined by the friction value, wherein thetarget trajectory is otherwise assessed to be unsuitable.